For example, the function in (a) is a parabola with an axis of symmetry at $x=1$ and a vertex at $(1,-9)$.

The function has $x$-intercepts at $x=-2$ and $x=4$, while it has an $y$-intercept at $y=-8$.

The function is negative on $-2<x<4$ and positive on $x4$.

The function is decreasing for $x1$.

Answers could vary. For example, the function in (a) is a parabola with an axis of symmetry at $x=1$ and a vertex at $(1,-9)$.

For example, the function in (a) is a parabola with an axis of symmetry at $x=1$ and a vertex at $(1,-9)$.

The function opens upwards.

The function has $x$-intercepts at $x=-2$ and $x=4$, while it has an $y$-intercept at $y=-8$.

The function is negative on $-2<x<4$ and positive on $x4$.

The function is decreasing for $x1$.

Answers could vary. For example, the function in (a) is a parabola with an axis of symmetry at $x=1$ and a vertex at $(1,-9)$.

a. Add 10 to both sides of the equation:

$$
2x=10
$$

Divide both sides of the equation by 2:

$$
x=5
$$

b. Subtract 6 from both sides of the equation:

$$
x=-6
$$

c. Zero product property:

$$
2x-10=0text{ or } x+6=0
$$

Use the result from (a) and (b):

$$
x=5text{ or }x=-6
$$

d. Subtract1 from both sides of the equation:

$$
4x=-1
$$

Divide both sides of the equation by 4:

$$
x=-dfrac{1}{4}
$$

e. Add 8 to both sides of the equation:

$$
x=8
$$

f. Zero product property:

$$
4x+1=0text{ or } x-8=0
$$

Use the result from (d) and (e):

$$
x=-dfrac{1}{4}text{ or }x=8
$$

a. $x=5$

b. $x=-6$

c. $x=5$ and $x=-6$

d. $x=-dfrac{1}{4}$

e. $x=8$

f. $x=-frac{1}{4}$ and $x=8$

Using the table,

Let
$$begin{aligned}
x&=text{figure number}\
y&=text{number of tiles}
end{aligned}$$

Based on the illustration, the **blue boxes** represent the **figure number multiply by itself**. The **yellow boxes** are represented by an expression $(2x-1)$.

Adding the two expressions.
$$begin{aligned}
y&=(x^2)+(2x-1)\
y&=x^2+2x-1enspacetext{(equation representing the number of tiles)}
end{aligned}$$

Using the equation in **Step 6** to find the number of tiles in Figure 5.
$$begin{aligned}
y&=x^2+2x-1
end{aligned}$$

Note that $x$ = figure number. Substituting it into the equation.
$$begin{aligned}
y&=(5)^2+2(5)-1\
y&=25+10-1\
y&=34enspacetext{tiles}
end{aligned}$$

Given that the perimeter of the figure is $76$ units, the sum of all the sides of the figure is equal to the perimeter. We determine the value of $x$ as follows:
$$begin{aligned}
P &= 76 \
P &= x + 2x + (x-1) + 2x + (2x-3) \
76 &= x + 2x + (x-1) + 2x + (2x-3) \
76 &= (x+2x +x+ 2x +2x) + (-1-3)\
76 &= 8x -4 \
76 + 4 &= 8x \
80 &= 8x \
x &= frac{80}{8} \
& boxed{x=10 text{ units}}
end{aligned}$$

Two sides of an isosceles triangle are congruent, hence, the length of the legs are equal. We calculate for the value of $x$ as follows:
$$begin{aligned}
L_1 &= L_2 \
3x-2 &= 5x – 14 \
3x-5x &= -14 + 2 \
-2x &= -12 \
x &= frac{-12}{-2} \
& boxed{x = 6}
end{aligned}$$

The figure shows supplementary angles, hence, sum of the angles is equal to $180^circ$. We determine the value of $x$ as follows:
$$begin{aligned}
theta_1 + theta_2 &= 180^circ \
3x + 20^circ + 5x &= 180^circ \
8x + 20^circ &= 180^circ \
8x &= 180^circ – 20^circ \
8x &= 160^circ \
x &= frac{160^circ}{8} \
& boxed{x = 20^circ}
end{aligned}$$

Since the two legs of an isosceles triangle are equal, the angles are also equal. We determine the value of $x$ as follows:
$$begin{aligned}
3x + 20^circ &= 5x \
3x -5x &= -20^circ \
-2x &= -20^circ \
x &= frac{-20^circ}{-2} \
& boxed{x=10^circ}
end{aligned}$$

a. $x = 10 text{ units}$
b. $x = 6$
c. $x=20^circ$
d. $x=10^circ$

Jackie made an error calculating the squares:

$$
(x+4)^2-2x-5=(x-1)^2
$$

Use $(apm b)^2=a^2pm 2ab +b^2$:

$$
x^2+8x+16-2x-5=x^2-2x+1
$$

Subtract $x^2$ from both sides of the equation:

$$
6x+11=-2x+1
$$

Add $2x$ to both sides of the equation:

$$
8x+11=1
$$

Subtract 11 from both sides of the equation:

$$
8x=-10
$$

Divide both sides of the equation by 8:

$$
x=-dfrac{5}{4}
$$

$$
x=-dfrac{5}{4}
$$

a. The sum of all angles in a triangle should be 180$text{textdegree}$:

$$
180text{textdegree}-47text{textdegree}-27text{textdegree}=106text{textdegree}
$$

The two triangles are then similar because of AA (two congruent angles).

b. Determine the ratios of the sides:

$$
dfrac{98}{18}approx 5.44
$$

$$
dfrac{64}{12}approx 5.33
$$

Since the ratios are not the same, the rectangles are not similar.

Yes, you can determine the length of each side using the distance formula or the Pythagorean theorem:

$$
SE=LI=MI=sqrt{3^2+1^2}=sqrt{10}
$$

$$
MS=LE=sqrt{3^2+2^2}=sqrt{13}
$$

Yes, $MS=LE$ and $SE=LI=MI$

a. Take the square root of both sides of the equation:

$$
x=pm 4
$$

b. Evaluate the square root:

$$
x=sqrt{16}=4
$$

c. If the absolute value of $x$ is 4, then $x$ can be $4$ or $-4$:

$$
x=pm 4
$$

d. Subtract 3 from both sides of the equation:

$$
x=sqrt{16}-3=4-3=1
$$

e. The square root of a number cannot be negative and thus no $x$ value will make this equation true.

f. Subtract 4 from both sides of the equation:

$$
x^2=-4
$$

The square root of a number cannot be negative and thus no $x$ value will make this equation true.

a. $x=pm 4$

b. $x=4$

c. $x=pm 4$

d. $x=1$

e. No solution

f. No solution

b. The probability is given in the row containing B and in the column containing B:

$$
dfrac{1}{4}=0.25=25%
$$

c. The probability is the sum of all probabilities that neither photo is of the niece:

$$
dfrac{1}{9}+dfrac{1}{6}+dfrac{1}{6}+dfrac{1}{4}=dfrac{4}{36}+dfrac{6}{36}+dfrac{6}{36}+dfrac{9}{36}=dfrac{25}{36}approx 0.694=69.4%
$$

a. Modle

b. $frac{1}{4}=0.25=25%$

c. $frac{25}{36}approx 0.694=69.4%$

The tangent ratio is the opposite side divided by the adjacent rectangular side:

$$
theta=tan^{-1}{dfrac{3}{4}}approx 37text{textdegree}
$$

37$text{textdegree}$

Corresponding sides of similar figures have the same proportions:

$$
dfrac{y}{3}=dfrac{15}{6}
$$

Multiply both sides of the equation by 3:

$$
y=dfrac{15cdot 3}{6}=7.5
$$

Determine $z$:

$$
dfrac{z}{24}=dfrac{6}{15}
$$

Multiply both sides of the equation by 24:

$$
z=dfrac{6cdot 24}{15}=9.6
$$

$y=7.5$ and $z=9.6$

The sum of all angles in a triangle is 180$text{textdegree}$:

$$
90+6x+8+2x+2=180
$$

Combine like terms:

$$
8x+100=180
$$

Subtract 100 from both sides of the equation:

$$
8x=80
$$

Divide both sides of the equation by 8:

$$
x=10
$$

Replace $x$ with 10 in the expression of the measure of the angles:

$$
mangle A=6x+8=6(10)+8=68text{textdegree}
$$

$$
mangle B=2x+2=2(10)+2=22text{textdegree}
$$

The angles are acute, since they are both less than 90$text{textdegree}$.

The angles are also complementary because their sum is 90$text{textdegree}$.

A line with slope $a$ and through point $(x_1,y_1)$ has general equation:

$$
y-y_1=a(x-x_1)
$$

a. Perpendicular lines have slopes whose product is $-1$, thus the perpendicular line then has slope $dfrac{5}{2}$:

$$
y+3=dfrac{5}{2}(x-2)
$$

b. Determine the slope of the given line:

$$
-3x+2y=10
$$

Add $3x$ to both sides of the equation:

$$
2y=10+3x
$$

Divide both sides of the equation by 2:

$$
y=5+dfrac{3}{2}x
$$

Thus the line has slope $dfrac{3}{2}$. Since parallel lines have the same slopes, the parallel line will have slope $dfrac{3}{2}$:

$$
y-7=dfrac{3}{2}(x-4)
$$

a. $y+3=dfrac{5}{2}(x-2)$

b. $y-7=dfrac{3}{2}(x-4)$

b. The $x$-intercepts are the intersections of the graph we the $x$-axis and thus are:

Maggie: $x=3$ and $x=14$

Jen: $x=0$ and $x=8$

Imp: $x=2$ and $x=12$

The difference between the $x$-intercepts is the distance that the balloon traveled.

c. The vertex is the maximum of each parabola:

Maggie: $(8.5,30.25)$

Jen: $(4,16)$

Imp: $(7,25)$

a. Maggie Nanimos won the Highest Launch and won the Longest distance.

b.

Maggie: $x=3$ and $x=14$

Jen: $x=0$ and $x=8$

Imp: $x=2$ and $x=12$

c.

Maggie: $(8.5,30.25)$

Jen: $(4,16)$

Imp: $(7,25)$

a. Replace $x$ in the equation with 2:

$$
y=-(2)^2+17(2)-42=-4+34-42=-12
$$

This tells you that the height would have been 12 yards under the ground. This is impossible and is caused by the fact that the balloon is launched further away at 3 yards.

b. An appropriate domain would then contain all $x$-values between the launch and the landing: $(3,14)$.

The range contains only non-negative distances: $[0, +infty)$.

a. Impossible

c. Domain $(3,14)$ and range $[0,+infty)$

a. Figure 1 contains 8 blocks, while figure 2 contains 14 blocks, figure 3 contains 22 blocks. Thus the number of blocks increases by 6 and then 8 blocks.

b. We note that the blocks form a rectangle with length $n+2$ and width $n+1$ that are increased by 2 blocks:

$$
f(n)=(n+2)(n+1)+2=n^2+3n+2+2=n^2+3n+4
$$

c. The equation is quadratic, because if simplified you obtain a term $n^2$ that is quadratic (and their are no terms with a greater power).

a. The number of blocks increases by 6 and then 8 blocks.

b. $f(x)=n^2+3n+4$

c. Yes

The graph is a parabola with symmetry axis $x=4$ and vertex $(4,-9)$.

The graph opens upward.

The graph is negative on $(1,7)$ and positive on $(-infty, 1)$ and $(1,+infty)$.

The function is decreasing on $(-infty,4)$ and increasing on $(4,+infty)$.

The graph is a parabola with symmetry axis $x=4$ and vertex $(4,-9)$.

**Part a**
We calculate the sum of the following expressions:
$$begin{aligned}
text{i. }& 0+3 = boxed{3} \
text{ii. }& -7 + 0 = boxed{-7} \
text{iii. }& 0 + 6 = boxed{6} \
text{iv. }& 0 + (-2) = boxed{-2} \
end{aligned}$$

**Part c**
When Julia added number $a$ and $b$ she was able to to get a sum of $b$, this means that either $a$ is equal to zero or $a$ and $b$ are both zero.

**Part d**
We calculate the product of the following expressions:
$$begin{aligned}
text{i. }& 3cdot 0 = boxed{0} \
text{ii. }& (-7) cdot 0 = boxed{0} \
text{iii. }& 0 cdot 6 = boxed{0} \
text{iv. }& 0 cdot (-2) = boxed{0} \
end{aligned}$$

**Part f**
When Julia multiplied the number $a$ and $b$, she was able to to get a product of $0$. This means that either one number is equal to zero or both are zero.

The sine ratio is the opposite side divided by the hypotenuse:

$$
sin{75text{textdegree}}=dfrac{26}{x}
$$

Multiply both sides of the equation by $x$:

$$
xsin{75text{textdegree}}=26
$$

Divide both sides of the equation by $sin{75text{textdegree}}$:

$$
x=dfrac{26}{sin{75text{textdegree}}}approx 27
$$

Thus the ladder needs to be about 27 ft.

a. We note that the temperature increases as the time increases, but also that the graph seems to level off (becomes constant) for $t>30$. Thus the corresponding graph is Graph 1.

b. The time increases as the distance increases and the time increases in an almost linear more, thus the group should be almost a line, this means that the graph is Graph 2.

a. The green triangle is the enlarged triangle and its coordinates are $A'(-3,-3), B'(9,-3)$ and $C'(-3,-6)$. We note that all original coordinates have been multiplied by the dilation factor.

b. The purple triangle is the rotated triangle and its coordinates are $A”(-3,3), B”(-3,-9)$ and $C”(-6,3)$. The $y$-coordinates have changed sign.

c. Since the $x$-coordinates of $A$ increased by 6 and the $y$ coordinates increased by 4, the same happens to the coordinates of $B$: $(3+6,-1+4)=(9,3)$

a. $A'(-3,-3)$, $B'(9,-3)$, $C'(-3,-6)$

b. $A”(-3,3)$, $B”(-3,-9)$, $C”(-6,3)$

c. $(9,3)$

A scale model is similar to the original model and corresponding measurements of similar objects have the same proportions.

$$
dfrac{x}{41}=dfrac{324}{128}
$$

Multiply both sides of the equation by 41:

$$
x=dfrac{324cdot 41}{128}=dfrac{13284}{128}=dfrac{3321}{32}approx 103.78m
$$

a. No, because you cannot determine how wide the function will open and for example you cannot determine the $x$-intercepts.

b. No, because for example you cannot know where the graph will intersect the $y$-axis.

c. Yes, in this case we can only draw exactly one parabola.

a. All $x$-intercepts have $y=0$ because the $x$-intercepts lie on the $x$-axis. All $y$-intercepts have $x=0$ because the $y$-intercepts lie on the $y$-axis.

b. Replace $x$ with 0 in the equation:

$$
y=2(0)^2+5(0)-12=-12
$$

c. Replace $y$ with 0:

$$
0=2x^2+5x-12
$$

Thus this would be the equation that would need to be solved.

d. Yes, Factorize:

$$
0=(2x-3)(x+4)
$$

Zero product property:

$$
2x-3=0text{ or }x+4=0
$$

Solve each equation to $x$:

$$
x=dfrac{3}{2}text{ or }x=-4
$$

a. $y=0$, $x=0$

b. $(0,-12)$

c. $0=2x^2+5x-12$

d. $x=frac{3}{2}$ or $x=-4$

a. Factorize:

$$
0=(2x-3)(x+4)
$$

b. Zero product property:

$$
2x-3=0text{ or }x+4=0
$$

Solve each equation to $x$:

$$
x=dfrac{3}{2}text{ or }x=-4
$$

c. You need to determine the $y$-intercept, which is the constant term of the equation of the parabola: $y=-12$.

a. $0=(2x-3)(x+4)$

b. $x=frac{3}{2}$ or $x=-4$

c. $y$-intercept $(0,-12)$

**Concept**
The zero product property states that if the product of $a$ and $b$ is zero, then $a=0$, $b=0$, or both numbers are equal to zero.

**Part a**
We determine the x-intercepts of the parabola $y=x^2+4x-5$ by setting the value of $y$ to zero. We calculate for the value of $x$ using the zero product property as follows:
$$begin{aligned}
y&=x^2+4x-5 \
0 &= x^2 +4x-5 \
0 &= (x+5)(x-1) \
x + 5 &= 0 \
& boxed{x = -5} \
x – 1 &= 0 \
& boxed{x = 1} \
end{aligned}$$

Therefore, the x-intercepts of the parabola are the points $boxed{(1,0)}$ and $boxed{(-5,0)}$

**Part b**
The vertex of a parabola is the point where the line of symmetry and the parabola intersects. The axis of symmetry of the parabola is between the x-intercepts, it is a line that passes through the midpoint of these intercepts. We determine the axis of symmetry as follows:
$$begin{aligned}
text{axis of symmetry} &= frac{x_1 + x_2}{2} \
text{axis of symmetry} &= frac{-5+1}{2} \
text{axis of symmetry} &= -2
end{aligned}$$

Therefore, the axis of symmetry is $x=-2$

Using the axis of symmetry at $x=-2$, we obtain the intersection of the axis of symmetry and the parabola. This intersection is the vertex. We determine the vertex using the graph as follows:

From the graph, we observe that the vertex of the parabola is the point $boxed{(-2,9)}$

**Part c**
The sketch of the parabola $y=x^2 + 4x -5$ along with its intercepts and vertex is as follows:

The $y$-intercept is the constant term of the equation and thus is $y=-6$.

Determine the $x$-intercepts. Replace $y$ with 0:

$$
0=x^2+x-6
$$

Factorize:

$$
0=(x+3)(x-2)
$$

Zero product property:

$$
x+3=0text{ or }x-2=0
$$

Solve each equation to $x$:

$$
x=-3text{ or }x=2
$$

Thus the $x$-intercepts are $x=-3$ and $x=2$.

The axis of symmetry then lies in the middle between the two $x$-intercepts and thus at $x=-0.5$. Determine the corresponding $y$-value:

$$
y=(-0.5)^2+(-0.5)-6=-6.25
$$

The vertex is then $(-0.5,-6.25)$.

Vertex $(-0.5,-6.25)$

$y$-intercept $(0,-6)$

$x$-intercepts $(-3,0)$ and $(2,0)$

**Concept**

The zero product property states that if the product of $a$ and $b$ is zero, then $a=0$, $b=0$, or both numbers are equal to zero.

**Part a**
The equations have similar factors, which are:
$$begin{equation}
(x+2)
end{equation}$$
$$begin{equation}
(x-1)
end{equation}$$
But they are using different operations. The first one is a multiplication operation, while the other one is an addition.

**Part b**
For the first equation:
Since the operation is multiplication, use Zero Product Property.
$$begin{aligned}
(x+2)(x-1) &=0 \
end{aligned}$$

becomes
$$begin{aligned}
(x+2) &=0 \
end{aligned}$$
and
$$begin{aligned}
(x-1) &=0 \
end{aligned}$$

Solve for $x$ on both factors.
$$begin{aligned}
x+2-2 &= 0-2 \
&boxed{x = -2} \
end{aligned}$$
and
$$begin{aligned}
x-1+1 &=0+1 \
&boxed{x =1} \
end{aligned}$$

For the second equation:
Simply add the factors
$$begin{aligned}
(x+2)+(x-1) &=0 \
x+2+x-1 &=0 \
2x+1 &= 0\
2x+1-1 &= 0-1\
2x &= -1\
frac{2x}{2} &= frac{-1}{2}\
&boxed{x = frac{-1}{2}}\
end{aligned}$$

a. They have similar factors but different operations.
b. The value of $x$ for:
Equation 1 are $x = -2$ and $x =1$
Equation 2 is $x = -frac{1}{2}$

**Concept**

The zero product property states that if the product of $a$ and $b$ is zero, then $a=0$, $b=0$, or both numbers are equal to zero.

**Part a**
Use Zero Product Property.
$$begin{aligned}
(x-2)(x+8) &=0 \
end{aligned}$$

becomes
$$begin{aligned}
(x-2) &=0 \
end{aligned}$$
and
$$begin{aligned}
(x+8) &=0 \
end{aligned}$$

Solve for $x$ on both factors.
$$begin{aligned}
x-2+2 &= 0+2 \
&boxed{x = 2} \
end{aligned}$$
and
$$begin{aligned}
x+8-8 &=0-8 \
&boxed{x = -8} \
end{aligned}$$

**Part b**
Use Zero Product Property.
$$begin{aligned}
(3x-9)(x-1) &=0 \
end{aligned}$$

becomes
$$begin{aligned}
(3x-9) &=0 \
end{aligned}$$
and
$$begin{aligned}
(x-1) &=0 \
end{aligned}$$

Solve for $x$ on both factors.
$$begin{aligned}
3x-9+9 &= 0+9 \
3x &= 9 \
frac{3x}{3} &= frac{9}{3} \
&boxed{x = 3} \
end{aligned}$$
and
$$begin{aligned}
x-1+1 &=0+1 \
&boxed{x = 1} \
end{aligned}$$

**Part c**
Use Zero Product Property.
$$begin{aligned}
(x+10)(2x-5) &=0 \
end{aligned}$$

becomes
$$begin{aligned}
(x+10) &=0 \
end{aligned}$$
and
$$begin{aligned}
(2x-5) &=0 \
end{aligned}$$

Solve for $x$ on both factors.
$$begin{aligned}
x+10-10 &=0-10\
&boxed{x = -10}
end{aligned}$$
and
$$begin{aligned}
2x-5 &=0 \
2x-5+5 &= 0+5 \
2x &= 5 \
frac{2x}{2} &= frac{5}{2} \
&boxed{x = frac{5}{2}} \
end{aligned}$$

**Part d**
Use Zero Product Property.
$$begin{aligned}
(x-7)^2 &=0 \
end{aligned}$$

becomes
$$begin{aligned}
(x-7) &=0 \
end{aligned}$$
and
$$begin{aligned}
(x-7) &=0 \
end{aligned}$$

Solve for $x$ on both factors.
$$begin{aligned}
x-7+7 &=0+7\
&boxed{x = 7}
end{aligned}$$
and
$$begin{aligned}
x-7+7 &=0+7\
&boxed{x = 7}
end{aligned}$$

a. $x = 2$ and $x = -8$
b. $x = 3$ and $x = 1$
c. $x = -10$ and $x = frac{5}{2}$
d. $x = 7$

**Concept**
The vertex formula can be computed as:
$$begin{aligned}
y=ax^2+bx+c
end{aligned}$$
Vertex $(h,k) = left(frac{-b}{2a}, frac{b^2-4ac}{4a} right)$

**Part a**
Compute for the x-coordinate of the vertex:
$$begin{aligned}
h &= frac{-b}{2a} \\
h &= frac{-(-16)}{2(2)} \\
&= frac{16}{4} \\
&boxed{= 4}
end{aligned}$$

**Part b**
To compute for the y-coordinate, substitute the obtained x-coordinate to the equation.
$$begin{aligned}
y &= 2x^2-16x+30 \
&= 2(4)^2-16(4)+30 \
&boxed{= -2}
end{aligned}$$

a. x-coordinate $=4$
b. y-coordinate $=-2$
The vertex is $(4,-2)$.

a. The probability is the height of the red bar:
$$
P(red)=dfrac{3}{8}=0.375=37.5%
$$

b. The probability is the height of the yellow bar:
$$
P(yellow)=dfrac{1}{8}=0.125=12.5%
$$

c. The probability is the height of the blue bar:
$$
P(blue)=dfrac{3}{8}=0.375=37.5%
$$

d. The total probability should be 1(=100%) and thus the remaining probability will be for the color green:

$$
P(green)=1-dfrac{3}{8}-dfrac{1}{8}-dfrac{3}{8}=dfrac{1}{8}=0.125=12.5%
$$

a. The monthly interest rate is the annual interest rate divided by the number of periods per year:

$$
dfrac{30%}{12}=2.5%=0.025
$$

b. The interest (which is what you will need to pay) is the product of the principal and the monthly interest rate (increased by 1) to the power of the number of months:

$$
f(x)=500(1.025)^x
$$

c. Replace $x$ with 6:

$$
f(6)=500(1.025)^6approx $579.85
$$

d. Replace $x$ with 12:

$$
f(12)=500(1.025)^{12}approx $672.44
$$

The interest is the new principal decreased by the previous principal:

$$
$672.44-$500=$172.44
$$

The effective annual rate of interest is then the interest divided by the original principal:

$$
dfrac{$172.44}{$500}=0.34488=34.488%
$$

a. 2.5%

b. $f(x)=500(1.025)^x$

c. $579.85

d. 34.488%

Vertical angles and alternate interior angles are congruent:

$$
10x+5=15x-30
$$

Subtract $10x$ from both sides of the equation:

$$
5=5x-30
$$

Add 30 to both sides of the equation:

$$
35=5x
$$

Divide both sides of the equation by 5:

$$
7=x
$$

$$
x=7text{textdegree}
$$

**Part a**
Solve the equation by completing the square method.
$$begin{aligned}
x^2+6x+8 &= 0 \
x^2+6x+8-8 &= 0-8 \
x^2+6x &= -8 \
end{aligned}$$

Get the coefficient of the $x$-term, divide it by $2$, and add its square to both sides of the equation.
$$begin{aligned}
x^2+6x+ left(frac{6}{2} right)^2 &= -8+left(frac{6}{2} right)^2 \
x^2+6x+ 9 &= 1 \
end{aligned}$$

Factor the equation.
$$begin{aligned}
x^2+6x+ 9 &= 1 \
(x+3)^2 &= 1 \
sqrt{(x+3)^2} &= sqrt{1} \
x+3 &= pm1 \
end{aligned}$$

Solve for $x$.
$$begin{aligned}
x+3 &= 1 \
x+3-3 &= 1-3 \
&boxed{x = -2} \
end{aligned}$$

and

$$begin{aligned}
x+3 &= -1 \
x+3-3 &= -1-3 \
&boxed{x = -4} \
end{aligned}$$

**Part b**
Solve the equation by factoring out $x$.
$$begin{aligned}
x^2+6x =0 \
x(x+6)=0
end{aligned}$$

Use Zero Product Property.
$$begin{aligned}
x(x+6)=0 \
end{aligned}$$
becomes
$$begin{aligned}
boxed{x=0} \
end{aligned}$$
and
$$begin{aligned}
x+6&=0 \
x+6-6&=0-6 \
&boxed{x=-6}
end{aligned}$$

**Part c**
Solve the equation by Zero Product Property.
$$begin{aligned}
0=3(x-5)(2x+3)\
end{aligned}$$
becomes
$$begin{aligned}
0&=x-5\

-x&=-5\
frac{-x}{-1} &= frac{-5}{-1}\
&boxed{x=5}\
end{aligned}$$

and

$$begin{aligned}
0&=2x+3\
-2x&=3\
frac{-2x}{-2} &= frac{3}{-2}\
&boxed{x=-frac{3}{2}}
end{aligned}$$

**Part d**
Solve the equation by completing the square method.
$$begin{aligned}
x^2 +4x-9 &= 3 \
x^2 +4x-9+9 &= 3+9 \
x^2 +4x &= 12 \
end{aligned}$$

Get the coefficient of the $x$-term, divide it by $2$, and add its square to both sides of the equation.
$$begin{aligned}
x^2+4x+ left(frac{4}{2} right)^2 &= 12+left(frac{4}{2} right)^2 \
x^2+4x+ 4 &= 16 \
end{aligned}$$

Factor the equation.
$$begin{aligned}
x^2+4x+ 4 &= 16 \
(x+2)^2 &= 16 \
sqrt{(x+2)^2} &= sqrt{16} \
x+2 &= pm 4 \
end{aligned}$$

Solve for $x$.
$$begin{aligned}
x+2 &= 4 \
x+2-2 &= 4-2 \
&boxed{x = 2} \
end{aligned}$$

and

$$begin{aligned}
x+2 &= -4 \
x+2-2 &= -4-2 \
&boxed{x = -6} \
end{aligned}$$

a. $x=-2$ and $x=-4$
b. $x=0$ and $x=-6$
c. $x=5$ and $x=-frac{3}{2}$
d. $x=2$ and $x=-6$

Based on the table in Step 2, the second difference has a constant rate of change. Therefore, the values in the table represents a quadratic equation. The standard form of a quadratic equation is,
$$begin{aligned}
y=ax^2+bx+c
end{aligned}$$

To get the value of $c$, the value of $x$ should be zero.
$$begin{aligned}
y&=a(0)^2+b(0)+c\
y&=c
end{aligned}$$

Based on the values in the table, when $x$ is zero, the value of $y$ is -6.
$$begin{aligned}
y&=c\
-6&=c
end{aligned}$$

Take the values of $x$ in table and substitute it in the quadratic equation found in **Step 3**.

When $x=1$,
$$begin{aligned}
y&=ax^2+bx+c\
y&=a(1)^2+b(1)+c\
y&=a+b+c
end{aligned}$$

When $x=2$,
$$begin{aligned}
y&=ax^2+bx+c\
y&=a(2)^2+b(2)+c\
y&=4a+2b+c
end{aligned}$$

When $x=3$,
$$begin{aligned}
y&=ax^2+bx+c\
y&=a(3)^2+b(3)+c\
y&=9a+3b+c
end{aligned}$$

When $x=4$,
$$begin{aligned}
y&=ax^2+bx+c\
y&=a(4)^2+b(4)+c\
y&=16a+4b+c
end{aligned}$$

Put the values of $x$ and $y$ in the table.

|x |0 |1 |2 | 3|4 |
|–|–|–|–|–|–|
|y |c |a+b+c |4a+2b+c |9a+3b+c |16a+4b+c |

Based on the illustration in **Step 12**, the values for $a$ and $b$ are,
$$begin{aligned}
2a&=2\
a&=dfrac{2}{2}\
a&=1
end{aligned}$$

$$begin{aligned}
a+b&=2\
1+b&=2\
b&=2-1\
b&=1
end{aligned}$$

Substituting the values of $a$, $b$, and $c$ into the standard form of quadratic equation.
$$begin{aligned}
y&=ax^2+bx+c\
y&=(1)x^2+(1)x-6\
y&=x^2+x-6
end{aligned}$$

Check the equation by selecting random points in the table.

Based on the table shown in **Step 12**, when $x$ = -4, the value of $y$ should be $6$,
$$begin{aligned}
y&=x^2+x-6\
y&=(-4)^2+(-4)-6\
y&=16-4-6\
y&=6
end{aligned}$$

Based on the table in **Step 17**, the second difference has a constant rate of change. Therefore, the values in the table represents a quadratic equation. The standard form of a quadratic equation is,
$$begin{aligned}
y=ax^2+bx+c
end{aligned}$$

Based on the equation in **Step 4**, when $x$ = 0,
$$begin{aligned}
y=c
end{aligned}$$

Based on the values in the table, when $x$ is zero, the value of $y$ is -6.
$$begin{aligned}
y&=c\
-5&=c
end{aligned}$$

The value of $y$ found in Steps **7-10** when $x=0$, $x=1$, $x=2$, $x=3$, and $x=4$ are the same.

Based on the illustration in **Step 23**, the values for $a$ and $b$ are,
$$begin{aligned}
2a&=2\
a&=dfrac{2}{2}\
a&=1
end{aligned}$$

$$begin{aligned}
a+b&=5\
1+b&=5\
b&=5-1\
b&=4
end{aligned}$$

Substituting the values of $a$, $b$, and $c$ into the standard form of quadratic equation.
$$begin{aligned}
y&=ax^2+bx+c\
y&=(1)x^2+(4)x-5\
y&=x^2+4x-5
end{aligned}$$

Check the equation by selecting random points in the table.

Based on the table shown in **Step 23**, when $x$ = -5, the value of $y$ should be $0$,
$$begin{aligned}
y&=x^2+4x-5\
y&=(-5)^2+4(-5)-5\
y&=25-20-5\
y&=0
end{aligned}$$

If $x=a$ is an $x$-intercept of the function (or a root), then the function equation has a factor $(x-a)$. In the table we note that $x=-3$ and $x=2$ is a root, thus the equation is then of the form:

$$
y=a(x+3)(x-2)
$$

Evaluate the function at one of the given points:

$$
-12=a(3)(-2)=-6a
$$

Divide both sides of the equation by $-6$:

$$
2=a
$$

Replace $a$ in the equation with 2:

$$
y=2(x+3)(x-2)
$$

$$
y=2(x+3)(x-2)
$$

If $(x_1, y_1)$ is the vertex of a parabola, then the equation of the parabola is of the form $f(x)=a(x-x_1)^2+y_1$.

We note that the vertex is $(7,25)$ (since in the table we note that ti is the maximum and $x=7$ is also obviously the axis of symmetry). The equation of the parabola is then of the form:

$$
y=a(x-7)^2+25
$$

Next evaluate the function at one point:

$$
0=a(2-7)^2+25=25a+25
$$

Subtract 25 from both sides of the equation:

$$
-25=25a
$$

Divide both sides of the equation by 25:

$$
-1=a
$$

Replace $y$ in the equation with $-1$:

$$
y=-(x-7)^2+25
$$

$$
y=-(x-7)^2+25
$$

If you are able to determine the vertex $(a,b)$, then the equation is of the form $f(x)=c(x-a)^2+b$.

If you are able to determine the $x$-intercepts from the table $x=a$ and $x=b$, then the equation is of the form $f(x)=c(x-a)(x-b)$.

Determine the remaining constant by evaluating the function at another point (than those you’ve used).

If $x=a$ is an $x$-intercept of the function (or a root), then the function equation has a factor $(x-a)$.

a. In the table we note that $x=3$ and $x=-5$ is a root, thus the equation is then of the form:

$$
y=a(x-3)(x+5)
$$

Evaluate the function at one of the given points:

$$
16=a(-3)(5)=-15a
$$

Divide both sides of the equation by $-15$:

$$
-dfrac{16}{15}=a
$$

Replace $a$ in the equation with $-dfrac{16}{15}$:

$$
y=-dfrac{16}{15}(x-3)(x+5)
$$

b. In the table we note that $x=1$ and $x=2.5$ is a root, thus the equation is then of the form:

$$
y=a(x-1)(x-2.5)
$$

Evaluate the function at one of the given points:

$$
-2.5=a(1)(2.5)=2.5a
$$

Divide both sides of the equation by $2.5$:

$$
-1=a
$$

Replace $a$ in the equation with $-1$:

$$
y=-(x-1)(x-2.5)
$$

c. In the table we note that $x=8$ is a root, thus the equation is then of the form:

$$
y=a(x-8)(x-b)
$$

Evaluate the function at two of the given points:

$$
-16=a(-8)(-b)=8abRightarrow a=dfrac{-16}{8b}=dfrac{-2}{b}
$$

$$
-2=a(-5)(3-b)=-5a(3-b)
$$

Replace $a$ in the second equation with $-dfrac{2}{b}$:

$$
-2=-dfrac{10}{b}(3-b)
$$

Use distributive property:

$$
-2=-dfrac{30}{b}+10
$$

Subtract 10 from both sides of the equation:

$$
-12=-dfrac{30}{b}
$$

Multiply both sides of the equation by $b$:

$$
-12b=-30
$$

Divide both sides of the equation by $-12$:

$$
b=2.5
$$

Determine $a$:

$$
a=dfrac{-2}{b}=dfrac{-2}{2.5}=-0.8
$$

Thus the equation then becomes:

$$
y=-0.8(x-8)(x-2.5)
$$

a. $y=-dfrac{16}{15}(x-3)(x+5)$

b. $y=-(x-1)(x-2.5)$

c. $y=-0.8(x-8)(x-2.5)$

a. Zero product property:

$$
6x-18=0text{ or }3x+2=0
$$

Solve each equation to $x$:

$$
6x=18text{ or }3x=-2
$$

$$
x=3text{ or }x=-dfrac{2}{3}
$$

b. Factorize:

$$
(x-5)(x-2)=0
$$

Zero product property:

$$
x-5=0text{ or }x-2=0
$$

Solve each equation to $x$:

$$
x=5text{ or }x=2
$$

c. Factorize:

$$
(x+4)(x-3)=0
$$

Zero product property:

$$
x+4=0text{ or }x-3=0
$$

Solve each equation to $x$:

$$
x=-4text{ or }x=3
$$

d. Factorize:

$$
(2x-1)(2x+1)=0
$$

Zero product property:

$$
2x-1=0text{ or }2x+1=0
$$

Solve each equation to $x$:

$$
2x=1text{ or }2x=-1
$$

$$
x=dfrac{1}{2}text{ or }x=-dfrac{1}{2}
$$

e. Subtract 9 from both sides of the equation:

$$
x^2-9=0
$$

Factorize:

$$
(x-3)(x+3)=0
$$

Zero product property:

$$
x-3=0text{ or }x+3=0
$$

Solve each equation to $x$:

$$
x=3text{ or }x=-3
$$

f. Factorize:

$$
(x-1)(x-1)=0
$$

Zero product property:

$$
x-1=0
$$

Add 1 to both sides of the equation:

$$
x=1
$$

a. $x=3$ or $x=-frac{2}{3}$

b. $x=5$ or $x=2$

c. $x=-4$ or $x=3$

d. $x=frac{1}{2}$ or $x=-frac{1}{2}$

e. $x=3$ or $x=-3$

f. $x=1$

a. Take the square root of both sides of the equation:

$$
x=pm 8
$$

b.Take the square root of both sides of the equation:

$$
x+1=pm 8
$$

Subtract 1 from both sides of the equation:

$$
x=-1pm 8=7text{ or }-9
$$

c. Add 64 to both sides of the equation:

$$
(x-1)^2=64
$$

Take the square root of both sides of the equation:

$$
x+1=pm 8
$$

Subtract 1 from both sides of the equation:

$$
x=-1pm 8=7text{ or }-9
$$

a. $x=pm 8$

b. $x=7$ or $x=-9$

c. $x=7$ or $x=-9$

The sine ratio is the opposite side divided by the hypotenuse:

$$
theta = sin^{-1}{dfrac{14}{16}}approx 61text{textdegree}
$$

61$text{textdegree}$

Determine $x$:

The sum of all angles in a triangle is 180$text{textdegree}$ and the base angles in a triangle are congruent:

$$
28+2x=180
$$

Subtract 28 from both sides of the equation:

$$
2x=152
$$

Divide both sides of the equation by 2:

$$
x=76
$$

Determine $y$:

Corresponding angles are equivalent since the lines are parallel:

$$
y=76(=x)
$$

Determine $z$:

Supplementary angles sum up to 180$text{textdegree}$:

$$
x+z=180
$$

Replace $x$ with its known value:

$$
76+z=180
$$

Subtract 76 from both sides of the equation:

$$
z=104
$$

$$
x=y=76text{textdegree}
$$

$$
z=104text{textdegree}
$$

Use the addition rule:

$$
P(preparationtext{ or }lost)=P(preparation)+P(lost)-P(preparationtext{ and }lost)
$$

The probability is the number of favorable outcomes divided by the number of possible outcomes:

$$
P(preparation)=dfrac{18}{200}=dfrac{9}{100}=0.09=9%
$$

$$
P(lost)=dfrac{12}{200}=dfrac{3}{50}=0.06=6%
$$

Fill in the known values in the equation:

$$
11%=9%+6%-P(preparationtext{ and }lost)
$$

Add $P(preparationtext{ and }lost)$ to both sides of the equation:

$$
11%+P(preparationtext{ and }lost)=9%+6%
$$

Subtract 11% from both sides of the equation:

$$
P(preparationtext{ and }lost)=9%+6%-11%=4%
$$

(1) The roots are $x=-7$ and $x=4$, thus the equation contains the factor $(x+7)$ and $(x-4)$. Multiply these factors:

$$
(x+7)(x-4)=x^2+7x-4x-28=x^2+3x-28
$$

Thus the equation is (b).

(2) The roots are $x=-6$ and $x=-4$, thus the equation contains the factor $(x+6)$ and $(x+4)$. Multiply these factors:

$$
(x+6)(x+4)=x^2+6x+4x+24=x^2+10x+24
$$

Thus the equation is (e).

(3) The double root is $x=-3$, thus the equation contains the factor $(x+3)^2$.
Thus the equation is (a).

(4) The roots are $x=0$ and $x=4$, thus the equation contains the factor $(x)$ and $(x-4)$. Multiply these factors:

$$
(x)(x-4)=x^2-4x
$$

Thus the equation is (g).

(5) The roots are $x=-2$ and $x=2$, thus the equation contains the factor $(x+2)$ and $(x-2)$. Multiply these factors:

$$
(x+2)(x-2)=x^2+2x-2x-4=x^2-4
$$

Note that the function also opens downwards, then we need to multiply the product of the factors with $-1$:

$$
(-1)(x^2-4)=-x^2+4
$$

Thus the equation is (d).

(6) The roots are $x=-dfrac{1}{2}$ and $x=dfrac{3}{2}$, thus the equation contains the factor $(x+0.5)$ and $(x-1.5)$. Multiply these factors:

$$
(x+0.5)(x-1.5)=x^2+0.5x-1.5x-0.75=x^2-x-0.75
$$

The factors can still be multiplied by a constant, then we note that equation (i) is the only multiple of these factors and thus this is the equation of the graph.

A. To be able to determine the equation of the parabola, we need to know at least one more point of the parabola (such that we have three points).

B. The $x$-intercepts are $x=-3$ and $x=2$. Then the equation is of the form:

$$
y=a(x+3)(x-2)
$$

Evaluate the function at one of the given points (note that $(0,6)$ is a point):

$$
6=a(3)(-2)=-6a
$$

Divide both sides of the equation by $-6$:

$$
-1=a
$$

Replace $a$ in the equation with $-1$:

$$
y=-(x+3)(x-2)
$$

C. We know that $x=5$ is a double root, then we know that the function is of the form:

$$
y=a(x-5)^2
$$

However if we want to determine the exact graph that she wants, we would need one more point, but we could try the graph with $a=1$ and see if this is what she wants (if she wants a more precise graph, she will need to supply an extra point).

$$
y=(x-5)^2
$$

A. need one more point

B. $y=-(x+3)(x-2)$

C. $y=(x-5)^2$

If $x=a$ is a root or $x$-intercept, then the function equation contains a factor of the form $(x-a)$.

a. The $x$-intercepts are $x=0$ and $x=8$. Then the equation is of the form:

$$
y=a(x)(x-8)
$$

Evaluate the function at one of the given points (note that $(4,32)$ is a point):

$$
32=a(4)(-4)=-16a
$$

Divide both sides of the equation by $-16$:

$$
-2=a
$$

Replace $a$ in the equation with $-2$:

$$
y=-2x(x-8)
$$

b. The $x$-intercepts are $x=10$ and $x=16$. Then the equation is of the form:

$$
y=a(x-10)(x-16)
$$

Evaluate the function at one of the given points (note that $(13,27)$ is a point, because of the symmetry of a parabola):

$$
27=a(3)(-3)=-9a
$$

Divide both sides of the equation by $-9$:

$$
-3=a
$$

Replace $a$ in the equation with $-3$:

$$
y=-3(x-10)(x-16)
$$

c. $d$ and $f$ are the $x$-intercepts (or also called the roots) of the graph, while $a$ can be determined by evaluating the function at one point (not one of the $x$-intercepts) and solving the equation to $a$.

a. $y=-2x(x-8)$

b. $y=-3(x-10)(x-16)$

c. See explanation

a. We know that $x=5$ is a double root since the graph hits the $x$-axis only once at $(5,0)$.

b. We know that if $x=a$ is a root, then $(x-a)$ is a factor of the function equation. Since $x=5$ is a double root, we know that $(x-5)^2$ will need to be a factor of the equation.

c. Rewriting the equation we obtain:
$$
y=x^2-8x+16=(x-4)^2
$$

We note that in this equation $x=4$ is a double root, however we needed a double root at $x=5$ and thus this function does not meet Vera’s needs.

a. $x=5$ is a double root

b. $(x-5)^2$ needs to be a factor of the equation

c. $x$-intercept and vertex is both $(4,0)$

**Part a**
Solve for $x$.
$$begin{aligned}
|x-4|=3
end{aligned}$$
becomes
$$begin{aligned}
x-4&=3\
x-4+4&=3+4\
&boxed{x=7}
end{aligned}$$
and
$$begin{aligned}
x-4+4&=-3+4 \
&boxed{x=1}
end{aligned}$$

Verify if $x$ makes the equation true.
For $x=7$:
$$begin{aligned}
|x-4| &stackrel{?}{=} 3\
|7-4| &stackrel{?}{=} 3\
3 &stackrel{checkmark}{=} 3\
end{aligned}$$
For $x=1$:
$$begin{aligned}
|1-4| &stackrel{?}{=} 3\
|-3| &stackrel{?}{=} 3\
3 &stackrel{checkmark}{=} 3\
end{aligned}$$

**Part b**
Solve for $x$.
$$begin{aligned}
|x-4|-3 &= 0\
|x-4|-3+3 &= 0+3\
|x-4| &= 3\
end{aligned}$$
becomes
$$begin{aligned}
x-4&=3\
x-4+4&=3+4\
&boxed{x=7}
end{aligned}$$
and
$$begin{aligned}
x-4+4&=-3+4 \
&boxed{x=1}
end{aligned}$$

Verify if $x$ makes the equation true.
For $x=7$:
$$begin{aligned}
|x-4|-3 &stackrel{?}{=} 0\
|7-4|-3 &stackrel{?}{=} 0\
|3|-3&stackrel{?}{=} 0\
0 &stackrel{checkmark}{=} 0\
end{aligned}$$
For $x=1$:
$$begin{aligned}
|x-4|-3 &stackrel{?}{=} 0\
|1-4|-3 &stackrel{?}{=} 0\
|-3|-3 &stackrel{?}{=} 0\
0 &stackrel{checkmark}{=} 0\
end{aligned}$$

**Part c**
Solve for $x$.
$$begin{aligned}
|x-4|=-3
end{aligned}$$
Since absolute value can never be negative, so there are **no solutions.**
$$begin{aligned}
boxed{x = empty}
end{aligned}$$

**Part d**
Solve for $x$.
$$begin{aligned}
(x-4)^2 &= 9\
sqrt{(x-4)^2} &= sqrt{9}\
x-4 &= pm 3\
end{aligned}$$

becomes

$$begin{aligned}
x-4 &= 3\
x-4+4&=3+4\
&boxed{x=7}
end{aligned}$$
and
$$begin{aligned}
x-4 &= -3\
x-4+4&=-3+4 \
&boxed{x=1}
end{aligned}$$

Verify if $x$ makes the equation true.
For $x=7$:
$$begin{aligned}
(x-4)^2 &stackrel{?}{=} 9\
(7-4)^2 &stackrel{?}{=} 9\
3^2 &stackrel{?}{=} 9\
9 &stackrel{checkmark}{=} 9\
end{aligned}$$
For $x=1$:
$$begin{aligned}
(x-4)^2 &stackrel{?}{=} 9\
(1-4)^2 &stackrel{?}{=} 9\
-3^2 &stackrel{?}{=} 9\
9 &stackrel{checkmark}{=} 9\
end{aligned}$$

**Part e**
Solve for $x$.
$$begin{aligned}
(x-4)^2-9 &= 0\
(x-4)^2-9+9 &= 0+9\
(x-4)^2 &= 9\
sqrt{(x-4)^2} &= sqrt{9}\
x-4 &= pm 3\
end{aligned}$$

becomes

$$begin{aligned}
x-4 &= 3\
x-4+4&=3+4\
&boxed{x=7}
end{aligned}$$
and
$$begin{aligned}
x-4 &= -3\
x-4+4&=-3+4 \
&boxed{x=1}
end{aligned}$$

Verify if $x$ makes the equation true.
For $x=7$:
$$begin{aligned}
(x-4)^2-9 &stackrel{?}{=} 0\
(7-4)^2-9 &stackrel{?}{=} 0\
3^2-9 &stackrel{?}{=} 0\
0 &stackrel{checkmark}{=} 0\
end{aligned}$$
For $x=1$:
$$begin{aligned}
(x-4)^2-9 &stackrel{?}{=} 0\
(1-4)^2-9 &stackrel{?}{=} 0\
-3^2-9 &stackrel{?}{=} 0\
0 &stackrel{checkmark}{=} 0\
end{aligned}$$

**Part f**
Solve for $x$.
$$begin{aligned}
(x-4)^2 &= -9\
end{aligned}$$
Since the equation is squared, the result can never be negative, so there are **no solutions.**
$$begin{aligned}
boxed{x = empty}
end{aligned}$$

a. $x=7$ and $x=1$
b. $x=7$ and $x=1$
c. $x = empty$
d. $x=7$ and $x=1$
e. $x=7$ and $x=1$
f. $x = empty$

Using a calculator, solve for the following:
**Part a**
$$begin{aligned}
&= 5-sqrt{36} \
&= 5-6 \
&boxed{= -1}
end{aligned}$$

**Part b**
$$begin{aligned}
&= 1+sqrt{39} \
&boxed{= 7.245}
end{aligned}$$

**Part c**
$$begin{aligned}
&= -2-sqrt{5} \
&boxed{= -4.236}
end{aligned}$$

a. $-1$
b. $7.245$
c. $-4.236$

a. The roots are $x=-4$ and $x=2$, thus the equation contains the factor $(x+4)$ and $(x-2)$. Thus the equation is of the form:

$$
y=a(x+4)(x-2)
$$

b. The double root is $x=3$, thus the equation contains the factor $(x-3)^2$. Thus the equation is of the form:

$$
y=a(x-3)^2
$$

c. The roots are $x=0$ and $x=7$, thus the equation contains the factor $(x)$ and $(x-7)$. Thus the equation is of the form:

$$
y=a(x)(x-7)
$$

d. The roots are $x=-5$ and $x=1$, thus the equation contains the factor $(x+5)$ and $(x-1)$. Thus the equation is of the form:

$$
y=a(x+5)(x-1)
$$

a. $y=a(x+4)(x-2)$

b. $y=a(x-3)^2$

c. $y=ax(x-7)$

d. $y=a(x+5)(x-1)$

The graph is a parabola with symmetry axis $x=1$ and vertex $(1,-9)$.

The graph opens upward.

The $x$-intercepts are $x=-2$ and $x=4$, while the $y$-intercept is $x=-8$.

The graph is negative on $(-2,4)$ and positive on $(-infty, -2)$ and $(4,+infty)$.

The function is decreasing on $(-infty,-2)$ and increasing on $(4,+infty)$.

The graph is a parabola with symmetry axis $x=1$ and vertex $(1,-9)$.

a. The tangent ratio is the opposite side divided by the adjacent rectangular side.

$$
theta=tan^{-1}dfrac{9}{3}approx 72text{textdegree}
$$

b. The equation of a line in slope-intercept form is $y=ax+b$ with $a$ the slope and $b$ the $y$-intercept.

The slope angle is 45$text{textdegree}$ if the slope is equal to:

$$
tan{45text{textdegree}}=1
$$

We have also given that $y=3$ is the $y$ intercept. Thus the equation is then:

$$
y=1x+3=x+3
$$

c. The given system of equations:

$$
y=3x+1
$$

$$
y=x+3
$$

Subtract the two equations:

$$
0=2x-2
$$

Add 2 to both sides of the equation:

$$
2=2x
$$

Divide both sides of the equation by 2:

$$
1=x
$$

Determine $y$:

$$
y=3x+1=3(1)+1=4
$$

Thus the solution is $(1,4)$.

a. 72$text{textdegree}$

b. $y=x+3$

c. $(1,4)$

a. The two triangles are congruent because of AAS, because the two bottom angles of the triangle are congruent (since the triangle is isosceles), we also know a pair of right angles and we know a pair of congruent sides (QS with itself)

$$
triangle SQPcong triangle SQR
$$

b. The two triangles are congruent because of ASA and because alternate interior angles are equivalent if the lines are parallel.

$$
triangle EDGcong triangle GFE
$$

c. The two triangles are congruent because of SSS:

$$
triangle ABC cong triangle DEF
$$

a. The area of a square is the length of one of the sides squares:

$$
(x+3)^2=169
$$

Take the square root of both sides of the equation:

$$
x+3=pm 13
$$

Subtract 3 from both sides of the equation:

$$
x=-3pm 13=10text{ or } -16
$$

Only a positive length will make sense:

$$
x=10
$$

b. The length of the side of the square is the square root of the area:

$$
sqrt{12}=2sqrt{3}ft
$$

The length of a side of the large square is then:

$$
2sqrt{3}ft+12’=2sqrt{3}ft+1ft=(2sqrt{3}+1)ft
$$

The area of the larger square is then the length of one side squared:

$$
(2sqrt{3}+1)^2=12+4sqrt{3}+1=13+4sqrt{3}text{ square ft }
$$

Determine the difference in area:

$$
13+4sqrt{3}-12=1+4sqrt{3}approx 8text{ square feet}
$$

a. If $x=-1$ and $x=3$, then $x-1=pm 2$ and thus $(x-1)^2=4$

b. If $x=1$, then $x-1=0$ and thus $(x-1)^2=0$

c. The equation cannot have any $x$-values that make the equation true, because the square can never take on negative values.

d. You first determine the square root of both sides of the equation and then solve this equation to $x$.

a. $x=-1$ and $x=3$

b. $x=1$

c. No solution

d. You first determine the square root of both sides of the equation and then solve this equation to $x$.

a. Take the square root of both sides of the equation:

$$
x-3=pm sqrt{12}=pm 2sqrt{3}
$$

Add 3 to both sides of the equation:

$$
x=3pm 2sqrt{3}
$$

b. We note that there are two solutions.

c.
$$
2+sqrt{3}approx 3.732….
$$

$$
2-sqrt{3}approx 0.2679….
$$

a. $x=3pm 2sqrt{3}$

b. 2

c. $2+sqrt{3}approx 3.732…$ and $2-sqrt{3}approx 0.2679….$

a. Take the square root of both sides of the equation:

$$
2x-3=pm 7
$$

Add 3 to both sides of the equation:

$$
2x=3pm 7
$$

Divide both sides of the equation by 2:

$$
x=dfrac{3pm 7}{2}=5text{ or }-2
$$

Thus there are two solutions.

c. Take the square root of both sides of the equation:

$$
x+4=pm sqrt{20}=pm2sqrt{5}
$$

Subtract 4 from both sides of the equation:

$$
x=-4pm 2sqrt{5}
$$

Thus there are 2 solutions

d. Take the square root of both sides of the equation:

$$
5-10x=0
$$

Subtract 5 from both sides of the equation:

$$
-10x=-5
$$

Divide both sides of the equation by $-10$:

$$
x=dfrac{1}{2}
$$

Thus there is 1 solution.

f. Subtract 5 from both sides of the equation:

$$
(x+11)^2=0
$$

Take the square root of both sides of the equation:

$$
x+11=0
$$

Subtract 11 from both sides of the equation:

$$
x=-11
$$

Thus there is 1 solution.

d. The number of intersections of the graph(s) with the $x$-axis is equal to the number of solutions found of the equations in the previous exercise.

The number of intersections of the graph(s) with the $x$-axis is equal to the number of solutions found of the equations in the previous exercise.

a. On the graph we note that the $x$-intercepts are between -7 and -6 and between -2 and -1.

b. Add7 to both sides of the equation:

$$
7=(x+4)^2
$$

Take the square root of both sides of the equation:

$$
pm sqrt{7}= x+4
$$

Subtract 4 from both sides of the equation:

$$
-4pm sqrt{7}=x
$$

Thus we note that the solutions are irrational numbers (because the root can only be approximated by a decimal).

c. On the graph we note that the vertex is $(-4,7)$, this answer is exact because if an equation is of the form $y=(x-a)^2+b$ then $(a,b)$ is the vertex of the parabola.

a. Between -7 and -6 and between -2 and -1

b. Irrational

c. $(-4,7)$

**Concept**
Triangle Inequality Theorem states that the length of a side of a triangle is less than the sum of the lengths of the two sides and is greater than the absolute value of their difference.
$$begin{aligned}
|overline{ MS} -overline{SL}|<overline{ML}<overline{ MS} +overline{SL}
end{aligned}$$

**Solution**
Solve for the possible lengths of side $overline{ML}$.
$$begin{aligned}
\
|overline{ MS} -overline{SL}|&<overline{ML}<overline{ MS} +overline{SL} \
|10-4|&<overline{ML}<10+4 \
6&<overline{ML}<14\
end{aligned}$$

$6<overline{ML}<14$

Write the expressions in simplified radical form.
**Part a**
$$begin{aligned}
&= sqrt{3} cdot sqrt{6} \
&= sqrt{3 cdot 6}\
&= sqrt{18}\
&= sqrt{9 cdot 2}\
&= sqrt{9} cdot sqrt{2} \
&boxed{= 3 sqrt{2}} \
end{aligned}$$

**Part b**
$$begin{aligned}
&= 2sqrt{27} cdot 5sqrt{3} \
&= 10sqrt{81} \
&= 10 cdot 9 \
&boxed{= 90} \
end{aligned}$$

**Part c**
$$begin{aligned}
&= sqrt{5 cdot 16} \
&= sqrt{5} cdot sqrt{16} \
&boxed{= 4sqrt{5}} \
end{aligned}$$

a. $3 sqrt{2}$
b. $90$
c. $4sqrt{5}$

a. The sine ratio is the opposite side divided by the hypotenuse:

$$
theta = sin^{-1}{dfrac{6}{9}}approx 42text{textdegree}
$$

b. The cosine ratio is the adjacent rectangular side divided by the hypotenuse:

$$
theta = cos^{-1}{dfrac{5}{7}}approx 44text{textdegree}
$$

c. The tangent ratio is the opposite side divided by the adjacent rectangular side:

$$
theta = tan^{-1}{dfrac{7}{9}}approx 38text{textdegree}
$$

a. 42$text{textdegree}$

b. 44$text{textdegree}$

c. 38$text{textdegree}$

a. The quadrilateral is a trapezium because $overleftrightarrow{WZ}$ is parallel with $overleftrightarrow{XY}$.

c. Determine the lengths of the sides using the distance formula:

$$
WZ=sqrt{(6-7)^2+(5-3)^2}=sqrt{5}
$$

$$
WX=sqrt{(6-3)^2+(5-4)^2}=sqrt{10}
$$

$$
XY=sqrt{(9-3)^2+(1-4)^2}=sqrt{45}=3sqrt{5}
$$

$$
YZ=sqrt{(9-5)^2+(1-6)^2}=sqrt{41}
$$

The perimeter is then the sum of all lengths:

$$
4sqrt{5}+sqrt{10}+sqrt{41}
$$

c. The given transformation is the reflection about the $y$-axis, which changes the sign of the $x$-coordinates.

d. The rotation interchanges the $x$-and $y$-coordinates, and then changes the sign $y$-coordinates.

$$
begin{align*}
&W”(7,-3)
\ &Z”(6,-5)
end{align*}
$$

Determine the slope of the line $W”Z”$

$$
begin{align*}
text{Slope}&=frac{y_2-y_1}{x_2-x_1}=frac{-5-(-3)}{6-7}=frac{-5+3}{6-7}=frac{-2}{-1}=2
end{align*}
$$

a. Trapezium

b. $4sqrt{5}+sqrt{10}+sqrt{41}$

c. $Y'(-9,1)$

d. 2

a. Subtract 1 from both sides of the equation:

$$
x^2+4x=-1
$$

Complete the square:

$$
x^2+4x+4=-1+4
$$

Factorize:

$$
(x+2)^2=3
$$

Take the square root of both sides of the equation:

$$
x+2=pm sqrt{3}
$$

Subtract 2 from both sides of the equation:

$$
x=-2pm sqrt{3}
$$

b. Take the square root of both sides of the equation:

$$
x+2=pm sqrt{3}
$$

Subtract 2 from both sides of the equation:

$$
x=-2pm sqrt{3}
$$

a. $x=-2pm sqrt{3}$

b. $x=-2pm sqrt{3}$

a. Add 4 to both sides of the equation:

$$
x^2+6x+9=4
$$

Factorize:

$$
(x+3)^2=4
$$

Take the square root of both sides of the equation:

$$
x+3=pm 2
$$

Subtract 3 from both sides of the equation:

$$
x=-3pm 2=-1text{ or }-5
$$

b. Add 6 to both sides of the equation:

$$
x^2+2x+1=6
$$

Factorize:

$$
(x+1)^2=6
$$

Take the square root of both sides of the equation:

$$
x+1=pm sqrt{6}
$$

Subtract 1 from both sides of the equation:

$$
x=-1pmsqrt{6}
$$

a. $x=-1$ and $x=-5$

b. $x=-1pmsqrt{6}$

a. Add 9 to both sides of the equation:

$$
x^2+4x+4=9
$$

Factorize:

$$
(x+2)^2=9
$$

Take the square root of both sides of the equation:

$$
x+2=pm 3
$$

Subtract 2 from both sides of the equation:

$$
x=-2pm 3=1text{ or }-5
$$

b. Add 10 to both sides of the equation:

$$
x^2-6x+9=10
$$

Factorize:

$$
(x-3)^2=10
$$

Take the square root of both sides of the equation:

$$
x-3=pm sqrt{10}
$$

Add 3 to both sides of the equation:

$$
x=3pm sqrt{10}
$$

c. Add 4 to both sides of the equation:

$$
x^2-2x+1=5
$$

Factorize:

$$
(x-1)^2=5
$$

Take the square root of both sides of the equation:

$$
x-1=pm sqrt{5}
$$

Add 1 to both sides of the equation:

$$
x=1pm sqrt{5}
$$

a. $x=1$ and $x=-5$

b. $x=3pm sqrt{10}$

c. $x=1pm sqrt{5}$

The goal for this problem is to rephrase the given equation into perfect square form using algebra tiles and find the value of $x$.

Given equation.
$$begin{aligned}
2x^2+12x+3=11
end{aligned}$$

Simplify the equation by subtracting 3 on both sides of the equation.
$$begin{aligned}
2x^2+12x+3-textcolor{#4257b2}{3}&=11-textcolor{#4257b2}{3}\
2x^2+12x&=8
end{aligned}$$

Factor out $2$ on left side of the equation.
$$begin{aligned}
2(x^2+6x)=8
end{aligned}$$

DIvide $2$ on both sides of the equation.
$$begin{aligned}
dfrac{2(x^2+6x)}{2}&=dfrac{8}{2}\
x^2+6x&=4
end{aligned}$$

Based on the illustration in **Step 9**, the perfect square of the given equation is,
$$begin{aligned}
(x+3)^2=13
end{aligned}$$

Solving for the value of $x$ by using square root on both sides of the equation.
$$begin{aligned}
sqrt{(x+3)^2}&=pmsqrt{13}\
x+3&=pmsqrt{13}\
end{aligned}$$

Subtract $3$ on both sides of the equation.
$$begin{aligned}
x+3-textcolor{#4257b2}{3}&=pmsqrt{13}-textcolor{#4257b2}{3}\
x&=pmsqrt{13}-3\
end{aligned}$$

Solving for two values of $x$.
$$begin{aligned}
x&=sqrt{13}-3\
x&=0.605
end{aligned}$$

$$begin{aligned}
x&=-sqrt{13}-3\
x&=-6.605
end{aligned}$$

You build a perfect square if the expression is of the form $x^2pm 2ax+a^2$:

a. Note that $a=5$ and thus $a^2=25$ is the number of unit tiles.

b. Note that $a=3$ and thus $a^2=9$ is the number of unit tiles.

c. Note that $a=11$ and thus $a^2=11^2=121$ is the number of unit tiles.

**Part a**
We can calculate the value of the expression: $frac{2+sqrt{16}}{3}$ by getting the value of the radical as follows:
$$begin{aligned}
&= frac{2+sqrt{16}}{3} \
&= frac{2+4}{3} \
&= frac{6}{3} \
&= boxed{2}
end{aligned}$$

**Part b**
We can calculate the value of the expression: $frac{-1+sqrt{49}}{-2}$ by getting the value of the radical as follows:
$$begin{aligned}
&= frac{-1+sqrt{49}}{-2} \
&= frac{-1+7}{-2} \
&= frac{6}{-2} \
&= boxed{-3}
end{aligned}$$

**Part c**
We can calculate the value of the expression: $frac{-10 – sqrt{5}}{2}$ as by separating the expression as follows:
$$begin{aligned}
&= frac{-10 – sqrt{5}}{2} \
&= frac{-10}{2} – frac{sqrt{5}}{2} \
&= -5 – frac{sqrt{5}}{2} \
&approx boxed{-6.12}
end{aligned}$$

a. $2$
b. $-3$
c. $-6.12$

a. On the graph we note that the $x$-intercepts are between -2 and -1 and between 0 and 1. Moreover they are about $x=-1.3$ and $x=0.3$.

b. You cannot use the zero product property, because you cannot factorize the given equation, you will first need to complete the square.

a. $x=-1.3$ and $x=0.3$

b. Not possible to factorize equation

**Concept**
The standard form of a parabola can be described by the following equation:
$$begin{gather}
(x-h)^2 = 4p(y-k)
end{gather}$$

where the vertex at the point $(h,k)$, focus at $(h,k+p)$, and the directrix at $y=k-p$

The illustration is what a parabola that intersects the x-axis once looks like. The parabola that intersects the origin has the equation $x^2 = 4y$. This can be illustrated as follows:


The illustration is what a parabola that intersects the y-axis looks like. One example in which a parabola intersects the y-axis is the parabola with the equation $x^2 = 4(y-4)$. This can be illustrated as follows:


$$
overline{PQ}cong overline{SR}
$$

$$
overline{PR}cong overline{SQ}
$$

$$
overline{QR}cong overline{QR}
$$

$$
Downarrow SSS
$$

$$
triangle PQRsim triangle SRQ
$$

$$
Downarrow
$$

$$
angle P cong angle S
$$

The expected value is the sum of the product of the possibilities and the probability of the possibilities:

$$
EV=5cdot dfrac{1}{2}+3cdot dfrac{1}{4}+2cdot dfrac{1}{4}=dfrac{10}{4}+dfrac{3}{4}+dfrac{2}{4}=dfrac{15}{4}=3.75
$$

The number of tickets she is expected the win is then the product of the number of games and the expected value:

$$
3cdot 3.75=11.25
$$

Thus it is expected that she wins more than 10 tickets and thus we can predict that she will win enough tickets.

a. Add 2 to both sides of the equation:

$$
x^2-6x+9=2
$$

Factorize:

$$
(x-3)^2=2
$$

Take the square root of both sides of the equation:

$$
x-3=pm sqrt{2}
$$

Add 3 to both sides of the equation:

$$
x=3pm sqrt{2}
$$

b. Factorize:

$$
(p+1)^2=0
$$

Take the square root of both sides of the equation:

$$
p+1=0
$$

Subtract 1 from both sides of the equation:

$$
p=-1
$$

c. Subtract 5 from both sides of the equation:

$$
x^2-4x+4=-5
$$

Factorize:

$$
(x-2)^2=-5
$$

The equation has no solutions because the square is never negative.

a. $x=3pm sqrt{2}$

b. $p=-1$

c. No solutions

**a**. Completing the square of the quadratic equation of $(x^2-6x+7)$.
$$begin{aligned}
x^2-6x+7&=0\
x^2-6x&=-7\
end{aligned}$$

Add $9$ on both sides of the equation.
$$begin{aligned}
x^2-6x+9&=-7+9\
x^2-6x+9&=2
end{aligned}$$

Based on the diagram, the perfect square of the given quadratic equation is.
$$begin{aligned}
(x-3)^2&=2
end{aligned}$$

**b**. Completing the square of the given quadratic equation which is,
$$begin{aligned}
p^2+2p+1
end{aligned}$$

Based on the diagram, the perfect square of the equation $(p^2+2p+1)$ is
$$begin{aligned}
(p+1)^2
end{aligned}$$

**c**. Completing the square of the quadratic equation of $(k^2-4k+9)$.
$$begin{aligned}
k^2-4k+9&=0\
k^2-4k&=-9\
end{aligned}$$

Add $4$ on both sides of the equation.
$$begin{aligned}
k^2-4k+4&=-9+4\
k^2-4k+4&=-5
end{aligned}$$

Based on the diagram, the perfect square of the given quadratic equation is.
$$begin{aligned}
(k-2)^2&=-5
end{aligned}$$

a. We note that the missing unit tiles is a square with length 2.5:

$$
(2.5)^2=6.25
$$

b. Thus we need to add 6.25 to both sides of the equation:

$$
x^2+5x+6.25=-2+6.25=4.25
$$

a. 6.25

b. $x^2+5x+6.25=4.25$

a. The coefficient of $x$ is twice the constant in the square and the constant term is the difference between the square of this coefficient and the constant term (of the first equation).

b. Since the coefficient of $x$ is 10, the constant in the parentheses is half of 10 (which is 5) and thus the expression in the parentheses is $x+5$. The dimensions of the square is then $x+5$ by $x+5$.

a. The coefficient of $x$ is twice the constant in the square and the constant term is the difference between the square of this coefficient and the constant term (of the first equation).

b. $x+5$, $x+5$ by $x+5$

a. Add 144 to both sides of the equation:

$$
w^2+28w+196=144
$$

Factorize:

$$
(x+14)^2=144
$$

Take the square root of both sides of the equation:

$$
x+14=pm 12
$$

Subtract 14 from both sides of the equation:

$$
x=-14pm 12=-2text{ or }-26
$$

b. Add 2.25 to both sides of the equation:

$$
x^2+5x+6.25=2.25
$$

Factorize:

$$
(x+2.5)^2=2.25
$$

Take the square root of both sides of the equation:

$$
x+2.5=pm 1.5
$$

Subtract 2.5 from both sides of the equation:

$$
x=-2.5pm 1.5=-1text{ or }-4
$$

c. Add 64 to both sides of the equation:

$$
k^2-16k+64=81
$$

Factorize:

$$
(k-8)^2=9
$$

Take the square root of both sides of the equation:

$$
k-8=pm 3
$$

Add 8 to both sides of the equation:

$$
k=8pm 3=11text{ or }5
$$

d. Add 15 to both sides of the equation:

$$
x^2-24x+144=15
$$

Factorize:

$$
(x-12)^2=15
$$

Take the square root of both sides of the equation:

$$
x-12=pm sqrt{15}
$$

Add 12 to both sides of the equation:

$$
x=12pm sqrt{15}
$$

a. $w=-2$ and $w=-26$

b. $x=-1$ and $x=-4$

c. $k=5$ and $k=11$

d. $x=12pm sqrt{15}$

Completing the square

Make sure that the constant term (on the side with the $x$’s) is the square of half the coefficient of $x$ (by adding or subtracting the same number of both sides of the equation).

Then factorize (rewrite the expression with the $x$-terms as a square).

Next take the square root of both sides of the equation and solve to the unknown variable.

Make sure that the constant term (on the side with the $x$’s) is the square of half the coefficient of $x$ (by adding or subtracting the same number of both sides of the equation).

Then factorize (rewrite the expression with the $x$-terms as a square).

Next take the square root of both sides of the equation and solve to the unknown variable.

a. Add 4 to both sides of the equation:

$$
x^2+4x+4=1
$$

Factorize:

$$
(x+2)^2=1
$$

Take the square root of both sides of the equation:

$$
x+2=pm 1
$$

Subtract 2 from both sides of the equation:

$$
x=-2pm 1=-1text{ or }-3
$$

b. Add 9 to both sides of the equation:

$$
x^2-8x+16=9
$$

Factorize:

$$
(x-4)^2=9
$$

Take the square root of both sides of the equation:

$$
x-4=pm 3
$$

Add 4 to both sides of the equation:

$$
x=4pm 3=7text{ or }1
$$

c. Add 8.25 to both sides of the equation:

$$
x^2+5x+6.25=8.25
$$

Factorize:

$$
(x+2.5)^2=8.25
$$

Take the square root of both sides of the equation:

$$
x+2.5=pm sqrt{8.25}=pm dfrac{sqrt{33}}{2}
$$

Subtract 2.5 from both sides of the equation:

$$
x=-2.5pm dfrac{sqrt{33}}{2}=dfrac{-5pmsqrt{33}}{2}
$$

a. $x=-1$ and $x=-3$

b. $x=1$ and $x=7$

c. $x=dfrac{-5pmsqrt{33}}{2}$

a. Add 49 to both sides of the equation:

$$
x^2+6x+9=49
$$

Factorize:

$$
(x+3)^2=49
$$

Take the square root of both sides of the equation:

$$
x+3=pm 7
$$

Subtract 3 from both sides of the equation:

$$
x=-3pm 7=4text{ or }-10
$$

b. Factorize:

$$
(2x-3)(x+8)=0
$$

Zero product property

$$
2x-3=0text{ or }x+8=0
$$

Solve each equation to $x$:

$$
2x=3text{ or }x=-8
$$

$$
x=dfrac{3}{2}text{ or }x=-8
$$

a. $x=4$ and $x=-10$

b. $x=dfrac{3}{2}$ and $x=-8$

a.
$$
dfrac{-6+sqrt{196}}{2}=dfrac{-6+14}{2}=dfrac{8}{2}=4
$$

b.
$$
dfrac{-6-sqrt{196}}{2}=dfrac{-6-14}{2}=dfrac{-20}{2}=-10
$$

c.
$$
dfrac{-13-sqrt{361}}{4}=dfrac{-13-19}{4}=dfrac{-32}{4}=-8
$$

d.
$$
dfrac{-13+sqrt{361}}{4}=dfrac{-13+19}{4}=dfrac{6}{4}=dfrac{3}{2}
$$

We note that the solutions to the equations in the previous exercise are equal to the expressions evaluated here.

a. 4

b. $-10$

c. $-8$

d. $frac{3}{2}$

a. Definition absolute value:

$$
9+3x=pm 39
$$

Subtract 9 from both sides of the equation:

$$
3x=-9pm 39
$$

Divide both sides of the equation by 3:

$$
x=-3pm 13=10text{ or }-16
$$

b. Definition absolute value:

$$
2x+1=pm 10
$$

Subtract 1 from both sides of the equation:

$$
2x=-1pm 10
$$

Divide both sides of the equation by 2:

$$
x=-dfrac{1}{2}pm 5=dfrac{9}{2}text{ or }-dfrac{11}{2}
$$

c. Definition absolute value:

$$
-3x+9=pm 10
$$

Subtract 9 from both sides of the equation:

$$
-3x=-9pm 10
$$

Divide both sides of the equation by $-3$:

$$
x=3pm dfrac{10}{3}=dfrac{19}{3}text{ or }-dfrac{1}{3}
$$

a. $x=10$ or $x=-16$

b. $x=frac{9}{2}$ or $x=-frac{11}{2}$

c. $x=frac{19}{3}$ or $x=-frac{1}{3}$

d. No solution

a. The tangent ratio is the opposite side divided by the adjacent rectangular side:

$$
dfrac{Delta y}{Delta x}=dfrac{-5}{6}
$$

b. The distance we can determine using the Pythagorean theorem:

$$
LD=sqrt{5^2+6^2}=sqrt{61}approx 7.81
$$

c. The coordinates of $M$ are:

$$
M=left( -2+dfrac{2}{3}cdot 6, 1-dfrac{2}{3}cdot 5right)=left( 2, -dfrac{7}{3}right)
$$

d. You can determine the lengths using the distance formula:

$$
sqrt{(x_2-x_1)^2+(y_2-y_1)^2}
$$

while you can determine the slope using the ratio:

$$
dfrac{y_2-y_1}{x_2-x_1}
$$

a. $-frac{5}{6}$

b. $7.81$

c. $left(2,-frac{7}{3}right)$

d. Use distance formula

The probability is the number of favorable outcomes divided by the number of possible outcomes.

a. We note that there is 1 scalene triangle out of 4:

$$
P(scalene)=dfrac{1}{4}=0.25=25%
$$

b. We note that there are 3 isosceles triangles out of 4:

$$
P(isosceles)=dfrac{3}{4}=0.75=75%
$$

c. We note that 2 triangles out of 4 have a side with length 6 cm:

$$
P(6cm)=dfrac{2}{4}=dfrac{1}{2}=0.5=50%
$$

b. Add 9.25 to both sides of the equation:

$$
x^2-3x+2.25=9.25
$$

Factorize:

$$
(x-1.5)^2=9.25
$$

Take the square root of both sides of the equation:

$$
x-1.5=pm sqrt{9.25}=pm dfrac{sqrt{37}}{2}
$$

Add 1.5 to both sides of the equation

$$
x=1.5pm dfrac{sqrt{37}}{2}=dfrac{3}{2}pm dfrac{sqrt{37}}{2}approx 4.54text{ or }-1.54
$$

c. On the graph we note that the roots are about $-1.5$ and $4.5$ which agrees with the results from (b).

a. Not factorizable

b. $dfrac{3}{2}pm dfrac{sqrt{37}}{2}$

c. Yes

a. $a$ is the coefficient in front of $x^2$ and thus is 1. $b$ is the coefficient in front of $x$ and thus is $-3$. $c$ is the constant term and thus is $-7$.

b. Substitute the found values in (a) in the given equations:

$$
x=dfrac{3pm sqrt{(-3)^2-4(1)(-7)}}{2(1)}=dfrac{3pm sqrt{37}}{2}=dfrac{3}{2}pm dfrac{sqrt{37}}{2}
$$

We note that the solution is the same as that in the previous exercise.

a. $a=1$, $b=-3$ and $c=-7$

b. $x=dfrac{3}{2}pm dfrac{sqrt{37}}{2}$

The quadratic formula is
$$
x=dfrac{-bpm sqrt{b^2-4ac}}{2a}
$$

a.
$$
x=dfrac{-7pm sqrt{7^2-4(3)(2)}}{2(3)}=dfrac{-7pm 5}{6}=-dfrac{1}{3}text{ or } -2
$$

b.
$$
x=dfrac{9pm sqrt{(-9)^2-4(2)(-35)}}{2(2)}=dfrac{9pm 19}{4}=7text{ or } -dfrac{5}{2}
$$

c.
$$
x=dfrac{-10pm sqrt{10^2-4(8)(3)}}{2(8)}=dfrac{-10pm 2}{16}=-dfrac{1}{2}text{ or } -dfrac{3}{4}
$$

d.
$$
x=dfrac{5pm sqrt{(-5)^2-4(1)(9)}}{2(1)}=dfrac{5pm sqrt{-11}}{2}=text{ does not exist}
$$

a. $x=-frac{1}{3}$ or $x=-2$

b. $x=7$ or $x=-frac{5}{2}$

c. $x=-frac{1}{2}$ or $x=-frac{3}{4}$

d. No solution

The quadratic formula is
$$
x=dfrac{-bpm sqrt{b^2-4ac}}{2a}
$$

QUADRATIC FORMULA METHOD
$$
x=dfrac{-bpm sqrt{b^2-4(1)(c)}}{2(1)}=dfrac{-bpm sqrt{b^2-4c}}{2}
$$

COMPLETE THE SQUARE METHOD

Subtract $c$ from both sides of the equation:

$$
x^2+bx=-c
$$

Add $left( dfrac{b}{2}right)^2$ to both sides of the equation:

$$
x^2+bx+left( dfrac{b}{2}right)^2=-c+left( dfrac{b}{2}right)^2
$$

Factorize:

$$
left( x+dfrac{b}{2}right)^2=dfrac{b^2}{4}-c=dfrac{b^2-4c}{4}
$$

Take the square root of both sides of the equation:

$$
x+dfrac{b}{2}=pm dfrac{sqrt{b^2-4c}}{2}
$$

Subtract $dfrac{b}{2}$ from both sides of the equation:

$$
x=-dfrac{b}{2}pm dfrac{sqrt{b^2-4c}}{2}=dfrac{-bpm sqrt{b^2-4c}}{2}
$$

Thus we note that both methods give the same solution.

Quadratic formula

The quadratic formula is
$$
x=dfrac{-bpm sqrt{b^2-4ac}}{2a}
$$

where $a$ is the coefficient of $x^2$, $b$ is the coefficient of $x$ and $c$ is the constant term.

Solve the equation by replacing $a$, $b$ and $c$ by their values and then simplify to obtain the solution.

The quadratic formula is
$$
x=dfrac{-bpm sqrt{b^2-4ac}}{2a}
$$

where $a$ is the coefficient of $x^2$, $b$ is the coefficient of $x$ and $c$ is the constant term.

Solve the equation by replacing $a$, $b$ and $c$ by their values and then simplify to obtain the solution.

The quadratic formula is
$$
x=dfrac{-bpm sqrt{b^2-4ac}}{2a}
$$

with $a$ the coefficient of $x^2$, $b$ the coefficient of $x$ and $c$ the constant term.

a.
$$
x=dfrac{13pm sqrt{(-13)^2-4(1)(42)}}{2(1)}=dfrac{13pm 1}{2}=7text{ or } 6
$$

b.
$$
x=dfrac{-10pm sqrt{(10)^2-4(3)(-8)}}{2(3)}=dfrac{-10pm 14}{6}=dfrac{2}{3}text{ or } -4
$$

c.
$$
x=dfrac{10pm sqrt{(-10)^2-4(2)(0)}}{2(2)}=dfrac{10pm 10}{4}=5text{ or } 0
$$

d.
$$
x=dfrac{-8pm sqrt{8^2-4(4)(-60)}}{2(4)}=dfrac{-8pm 32}{8}=3text{ or }-5
$$

a. $x=7$ or $x=6$

b. $x=frac{2}{3}$ or $x=-4$

c. $x=5$ or $x=0$

d. $x=3$ or $x=-5$

a. We cannot apply the zero product property because the equation does not equal zero.

b. Use distributive property:

$$
x^2-5x+2x-10=-6
$$

Add 6 to both sides of the equation:

$$
x^2-3x-4=0
$$

Factorize:

$$
(x-4)(x+1)=0
$$

c. Zero product property:

$$
x-4=0text{ or }x+1=0
$$

Solve each equation to $x$:

$$
x=4text{ or }x=-1
$$

a. We cannot apply the zero product property because the equation does not equal zero.

b. $(x-4)(x+1)=0$

c. $x=4$ or $x=-1$

a. On the graph we note that the $x$-intercepts are $x=-3$ and $x=1$, thus the equation then contains the factors $(x+3)$ and $(x-1)$, thus the equation is of the form:

$$
y=a(x+3)(x-1)
$$

Another point on the graph is $(0,-3)$:

$$
-3=a(3)(-1)=-3a
$$

Divide both sides of the equation by $-3$:

$$
1=a
$$

Replace $a$ with 1 in the equation:

$$
y=(x+3)(x-1)
$$

b. In the table we note that the $x$-intercepts are $x=-2$ and $x=2$, thus the equation then contains the factors $(x+2)$ and $(x-2)$, thus the equation is of the form:

$$
y=a(x+2)(x-2)
$$

Another point on the graph is $(0,-4)$:

$$
-4=a(2)(-2)=-4a
$$

Divide both sides of the equation by $-4$:

$$
1=a
$$

Replace $a$ with 1 in the equation:

$$
y=(x+2)(x-2)
$$

a. $y=(x+3)(x-1)$

b. $y=(x+2)(x-2)$

c. Neither

The probability is the number of favorable outcomes divided by the number of possible outcomes:

$$
P(oak )=dfrac{35}{45}=dfrac{7}{9}
$$

The probability of the two events is then the product of the probability of both events (because they have been randomly assigned:

$$
dfrac{7}{9}cdot 20%=dfrac{7}{9}cdot dfrac{1}{5}=dfrac{7}{45}approx 0.156=15.6%
$$

a. Use the Pythagorean theorem:

$$
sqrt{4^2+3^2}=sqrt{25}=5
$$

$$
x=sqrt{5^2+12^2}=sqrt{169}=13
$$

b. Alternate interior angles are equivalent and supplementary angles sum up to 180$text{textdegree}$:

$$
x=180text{textdegree}-35text{textdegree}-65text{textdegree}=80text{textdegree}
$$

a. $x=13$

b. $x=80text{textdegree}$

a. Use the Pythagorean theorem:

$$
L:sqrt{6^2-3^2}=sqrt{27}=3sqrt{3}
$$

$$
R: sqrt{9^2-3^2}=sqrt{72}=6sqrt{2}
$$

The length is then the sum of the left and the right part:

$$
3sqrt{3}+6sqrt{2}
$$

b. The area of a triangle is the product of the base and the height, divided by 2

$$
AREA =dfrac{(3sqrt{3}+6sqrt{2})3}{2}=dfrac{9}{2}sqrt{3}+9sqrt{2}
$$

c. The perimeter is the sum of the lengths of all sides:

$$
PERIMETER=3sqrt{3}+6sqrt{2}+6+9=15+3sqrt{3}+6sqrt{2}
$$

a. $3sqrt{3}+6sqrt{2}$

b. $frac{9}{2}sqrt{3}+9sqrt{2}$

c. $15+3sqrt{3}+6sqrt{2}$

a. Factorize:

$$
(x+3)(x+9)=0
$$

Zero product property:

$$
x+3=0text{ or }x+9=0
$$

Solve each equation to $x$:

$$
x=-3text{ or }x=-9
$$

b. Factorize:

$$
(2x-11)^2=0
$$

Zero product property:

$$
2x-11=0
$$

Add 11 to both sides of the equation:

$$
2x=11
$$

Divide both sides of the equation by 2:

$$
x=dfrac{11}{2}
$$

b. Zero product property:

$$
3x+4=0text{ or }2x-1=0
$$

Solve each equation to $x$:

$$
3x=-4text{ or }2x=1
$$

$$
x=-dfrac{4}{3}text{ or }x=dfrac{1}{2}
$$

d. Subtract $8x-4$ from both sides of the equation:

$$
x^2-8x+16=0
$$

Factorize:

$$
(x-4)^2=0
$$

Zero product property:

$$
x-4=0
$$

Add 4 to both sides of the equation:

$$
x=4
$$

e. Subtract $60x$ from both sides of the equation:

$$
36x^2-60x+25=0
$$

Factorize:

$$
(6x-5)^2=0
$$

Zero product property:

$$
6x-5=0
$$

Add 5 to both sides of the equation:

$$
6x=5
$$

Divide both sides of the equation by 6:

$$
x=dfrac{5}{6}
$$

f. Subtract $2x+45$ from both sides of the equation:

$$
20x^2-32x-45=0
$$

Determine the roots using the quadratic formula:

$$
x=dfrac{32pm sqrt{(-32)^2-4(20)(-45)}}{2(20)}=dfrac{32pm 68}{40}=dfrac{5}{2}text{ or }-dfrac{9}{10}
$$

a. $x=-3$ or $x=-9$

b. $x=frac{11}{2}$

c. $x=-frac{4}{3}$ or $x=frac{1}{2}$

d. $x=4$

e. $x=frac{5}{6}$

f. $x=frac{5}{2}$ or $x=-frac{9}{10}$

Choosing a strategy to solve quadratic equations.

If the equation is easy to factorize or has been given in factorized form, then use the zero product property.

If the coefficient of $x^2$ is 1 and the coefficient of $x$ is even, then use the completing the square method.

Else use the quadratic formula.

If the equation is easy to factorize or has been given in factorized form, then use the zero product property.

If the coefficient of $x^2$ is 1 and the coefficient of $x$ is even, then use the completing the square method.

Else use the quadratic formula.

Let $x$ be the width of the enclosure, then $100-2x$ is the length of the enclosure. The area is then the product of the length and the width:

$$
x(100-2x)=600
$$

Subtract 600 from both sides of the equation:

$$
-2x^2+100x-600=0
$$

Determine the roots using the quadratic formula:

$$
x=dfrac{-100pm sqrt{100^2-4(-2)(-600)}}{2(-2)}=dfrac{-100pm sqrt{5200}}{-4}approx 6.97text{ or } 43.03
$$

Determine the length:

$$
100-2(6.97)approx 86.06
$$

$$
100-2(43.03)approx 13.94
$$

Since the length should be longer than the width, the length is then 86.06 ft and the width is 6.97 ft.

Let $x$ be the speed of Karen, then $10+x$ is the speed of Melinda. Since the distance is the product of the speed and the time:

$$
x+10+x=108
$$

Subtract 10 from both sides of the equation:

$$
2x=98
$$

Divide both sides of the equation by 2:

$$
x=49
$$

Detemrine the speed of Melinda:

$$
x+10=49+10=59
$$

Thus Karen drives 49 mph and Melinda drives 59 mph.

a. If $x=a$ is an $x$-intercept of the function (or a root), then the function equation has a factor $(x-a)$.

In the table we note that $x=-315$ and $x=315$ is a root, thus the equation is then of the form:

$$
y=a(x-315)(x+315)
$$

Evaluate the function at one of the given points ($(0,630)$ is a point of the parabola):

$$
630=a(-315)(315)=-99225a
$$

Divide both sides of the equation by $-99225$:

$$
-dfrac{2}{315}=a
$$

Replace $a$ in the equation with $-dfrac{2}{315}$:

$$
y=-dfrac{2}{315}(x-315)(x+315)
$$

b. Replace $x$ wth $-300$:

$$
y=-dfrac{2}{315}(-300-315)(-300+315)=dfrac{410}{7}approx 58.6ft
$$

c. Since the height cannot be negative, a useful domain is $[-315,315]$.

a. $y=-dfrac{2}{315}(x-315)(x+315)$

b. 58.6 ft

c. $[-315,315]$

a. Replace $y$ with 0:

$$
0=2x^2-4.8x
$$

Factorize:

$$
0=2x(x-2.4)
$$

Zero product property:

$$
x=0text{ or }x-2.4=0
$$

Solve each equation to $x$:

$$
x=0text{ or }x=2.4
$$

Thus the yo-yo was 2.4 s out of his hand.

b. Since a parabola is symmetry, it will take exactly half the time that the yo-yo was out of his hand:

$$
dfrac{2.4}{2}=1.2s
$$

c. Replace $x$ in the equation with 1.2:

$$
y=2(1.2)^2-4.8(1.2)=-2.88
$$

Thus the string is 2.88ft.

d.

The quadratic formula is
$$
x=dfrac{-bpm sqrt{b^2-4ac}}{2a}
$$

QUADRATIC FORMULA METHOD
$$
x=dfrac{-6pm sqrt{6^2-4(1)(11)}}{2(1)}=dfrac{-6pm sqrt{-8}}{2}=text{ does not exist }
$$

COMPLETE THE SQUARE METHOD

Subtract 2 from both sides of the equation:

$$
x^2+6x+9=-2
$$

Factorize:

$$
left( x+3right)^2=-2
$$

The equation has no solutions since the square is never negative.

a. Take the square root of both sides of the equation:

$$
x=pm 12
$$

b. Evaluate the square root:

$$
x=12
$$

c. Add 5 to both sides of the equation:

$$
x^2=144
$$

Take the square root of both sides of the equation:

$$
x=pm 12
$$

d. Evaluate the square root:

$$
x-1=12
$$

Add 1 to both sides of the equation:

$$
x=13
$$

f. Subtract 5 from both sides of the equation:

$$
x^2=-5
$$

The equation has no solutions, because the square is never negative.

a. $x=pm 12$

b. $x=12$

c. $x=pm 12$

d. $x=13$

e. No solutions

f. No solutions

a. Corresponding sides of similar figures have the same proportions:

$$
dfrac{x}{12}=dfrac{9}{6}
$$

Multiply both sides of the equation by 12:

$$
x=dfrac{9cdot 12}{6}=18
$$

b. Corresponding sides of similar figures have the same proportions:

$$
dfrac{w}{5}=dfrac{12}{3}
$$

Multiply both sides of the equation by 5:

$$
w=dfrac{12cdot 5}{3}=20
$$

c. Corresponding sides of similar figures have the same proportions:

$$
dfrac{n}{3}=dfrac{16}{7}
$$

Multiply both sides of the equation by 3:

$$
n=dfrac{16cdot 3}{7}=dfrac{48}{7}approx 6.86
$$

d. Corresponding sides of similar figures have the same proportions:

$$
dfrac{m}{7}=dfrac{11}{10}
$$

Multiply both sides of the equation by 7:

$$
m=dfrac{7cdot 11}{10}=dfrac{77}{10}=7.7
$$

a. $x=18$

b. $w=20$

c. $n=6.86$

d. $m=7.7$

The general equation of a line through two points $(x_1,y_1)$ and $(x_2,y_2)$ is
$$
y-y_1=dfrac{y_2-y_1}{x_2-x_1}(x-x_1)
$$

a. The line goes through the points $(0,3)$ and $(5,0)$:

$$
y-3=dfrac{0-3}{5-0}(x-0)
$$

Thus the equation is then:

$$
y=-dfrac{3}{5}x+3
$$

b. The $x$-intercept is the intersection of the graph with the $x$-axis and thus is $x=5$. The $y$-intercept is the intersection of the graph with the $y$-axis and thus is $y=3$.

c. The area of a triangle is the product of the base and the height divided by 2:

$$
dfrac{5cdot 3}{2}=7.5
$$

The perimeter is the sum of the lengths of all sides:

$$
3+5+sqrt{3^2+5^2}=8+sqrt{34}
$$

d. Perpendicular lines have slopes whose product is $-1$ and thus the slope of the perpendicular line is $dfrac{5}{3}$:

$$
y-3=dfrac{5}{3}(x-0)
$$

Thus the equation becomes:

$$
y=dfrac{5}{3}x+3
$$

a. $y=-frac{3}{5}x+3$

b. $x$-intercept $(5,0)$ and $y$-intercept $(0,3)$

c. Area 7.5 and Perimeter $8+sqrt{34}$

d. $y=frac{5}{3}x+3$

The tangent ratio is the opposite side divided by the adjacent rectangular side.

$$
tan 15.5text{textdegree}=dfrac{x}{5280}
$$

Multiply both sides of the equation by 5280:

$$
1,464approx 5280tan 15.5text{textdegree}=x
$$

Add the height of the eyes to the result:

$$
1464+5=1469
$$

Thus the heigh of the building is about 1,469 feet.

a.
$$
sqrt[3]{-64}=-sqrt[3]{64}=-sqrt[3]{4^3}=-4
$$

b.
$$
sqrt[5]{32}=sqrt[5]{2^5}=2
$$

c.
$$
sqrt[3]{-8}=-sqrt[3]{8}=-sqrt[3]{2^3}=-2
$$

d.
$$
sqrt[4]{10000}=sqrt[4]{10^4}=10
$$

a. $-4$

b. 2

c. $-2$

d. 10

a. Factorize:

$$
(100x-8)(100x+8)=0
$$

Zero product property:

$$
100x-8=0text{ or }100x+8=0
$$

Solve each equation to $x$:

$$
100x=8text{ or }100x=-8
$$

$$
x=dfrac{2}{25}text{ or }x=-dfrac{2}{25}
$$

b. Add $34x$ to both sides of the equation:

$$
9x^2+34x-8=0
$$

Factorize:

$$
(9x-2)(x+4)=0
$$

Zero product property:

$$
9x-2=0text{ or }x+4=0
$$

Solve each equation to $x$:

$$
9x=2text{ or }x=-4
$$

$$
x=dfrac{2}{9}text{ or }x=-4
$$

c. Determine the root using the quadratic formula:

$$
x=dfrac{4pm sqrt{(-4)^2-4(2)(7)}}{2(2)}=dfrac{4pm sqrt{-40}}{4}=text{ does not exist }
$$

d. Determine the root using the quadratic formula:

$$
x=dfrac{-0.2pm sqrt{(0.2)^2-4(3.2)(-5)}}{2(3.2)}=dfrac{-0.2pm sqrt{64.04}}{6.4}
$$

a. $x=frac{2}{25}$ or $x=-frac{2}{25}$

b. $x=frac{2}{9}$ or $x=-4$

c. No solution

d. $x=frac{-0.2pm sqrt{64.04}}{6.4}$

Corresponding angles of parallel lines are congruent:

$$
alpha = 85text{textdegree}
$$

Supplementary angles have sum 180$text{textdegree}$ and alternate interior angles of parallel lines are congruent:

$$
theta=180text{textdegree}-112text{textdegree}=68text{textdegree}
$$

$$
alpha = 85text{textdegree}
$$

$$
theta=68text{textdegree}
$$

a. Property absolute value:

$$
4x+20=pm 8
$$

Subtract 20 from both sides of the equation:

$$
4x=-20pm 8
$$

Divide both sides of the equatino by 4:

$$
x=-5pm 2=-3text{ or }-7
$$

b. Take the cubic root of both sides of the equation:

$$
x-13=2
$$

Add 13 to both sides of the equation:

$$
x=15
$$

c. Divide both sides of the equation by 2:

$$
sqrt{x-4}=7
$$

Square both sides of the equation:

$$
x-4=49
$$

Add 4 to both sides of the equation:

$$
x=53
$$

d. Add 4 to both sides of the equation:

$$
6|x-8|=18
$$

Divide both sides of the equation by 6:

$$
|x-8|=3
$$

Property absolute value:

$$
x-8=pm 3
$$

Add 8 to both sides of the equation:

$$
x=8pm 3=11text{ or } 5
$$

e. Divide both sides of the equation by 3:

$$
(x+12)^2=9
$$

Take the square root of both sides of the equation:

$$
x+12=pm 3
$$

Subtract 12 from both sides of the equation:

$$
x=-12pm 3=-9text{ or }-15
$$

f. Rewrite 36 as a power of 6:

$$
6^4=(6^2)^x=6^{2x}
$$

The powers have to be equal:

$$
4=2x
$$

Divide both sides of the equation by 2:

$$
2=x
$$

a. $x=-3$ or $x=-7$

b. $x=15$

c. $x=53$

d. $x=11$ or $x=5$

e. $x=-9$ or $x=-15$

f. $x=2$

a. Let $x$ be the speed of the walk. The distance traveled is the product of the speed and the time:

$$
(x+2)60=300
$$

Divide both sides of the equation by 60:

$$
x+2=5
$$

Subtract 2 from both sides of the equation:

$$
x=3
$$

Thus he needs to was 3 feet per second.

b. He will travel in the direction of the conveyor belt (because he walks slower than the conveyor belt). The distance traveled is the product of the speed and the time:

$$
(2-1)18=18
$$

Thus he travels for 18ft.

a. Replace $y$ with 0:

$$
0=3x^2-7x+4
$$

Factorize:

$$
0=(x-1)(3x-4)
$$

Zero product property:

$$
x-1=0text{ or }3x-4=0
$$

Solve each equation to $x$:

$$
x=1text{ or }3x=4
$$

$$
x=1text{ or }x=dfrac{4}{3}
$$

b. Replace $y$ with 0:

$$
(x+5)(-2x+3)=0
$$

Zero product property:

$$
x+5=0text{ or }-2x+3=0
$$

Solve each equation to $x$:

$$
x=-5text{ or }-2x=-3
$$

$$
x=-5text{ or }x=dfrac{3}{2}
$$

c. Replace $y$ with 0:

$$
x^2+6x=0
$$

Factorize:

$$
0=x(x+6)
$$

Zero product property:

$$
x=0text{ or }x+6=0
$$

Solve each equation to $x$:

$$
x=0text{ or }x=-6
$$

d. Replace $y$ with 0:

$$
0=3(x-5)(2x+3)
$$

Zero product property:

$$
x-5=0text{ or }2x+3=0
$$

Solve each equation to $x$:

$$
x=5text{ or }2x=-3
$$

$$
x=5text{ or }x=-dfrac{3}{2}
$$

a. $x=1$ or $x=frac{4}{3}$

b. $x=-5$ or $x=frac{3}{2}$

c. $x=0$ or $x=-6$

d. $x=5$ or $x=-frac{3}{2}$

b. False, for example the area of a cicle is $A=pi r^2$ with $r$ the radius of the circle.

c. True, because a 360$text{textdegree}$ rotation will rotate the figure onto itself.

a. 2, because the square equals to a positive number.

b. The graph will then have two $x$-intercepts, because you obtain the equation from (a) by letting $f(x)=0$ and add 2 to both sides of the equation.

c. Take the square root of both sides of the equation:

$$
x=pm sqrt{2}=pm 1.4142
$$

a. 2

b. 2

c. $x=pm sqrt{2}approx pm 1.4142$

You start by subtracting 1 from both sides of the equation:

$$
x^2=-1
$$

If you then take the square root of both sides of the equation, you obtain:

$$
x=pm sqrt{-1}
$$

But the square root of a negative number does not exist.

Use $i^2=-1$

a.
$$
sqrt{-4}=sqrt{4(-1)}=sqrt{4i^2}=2i
$$

b.
$$
(2i)(3i)=6i^2=6(-1)=-6
$$

c.
$$
-5(2i)^2=-5(4i^2)=-20i^2=-20(-1)=20
$$

d.
$$
sqrt{-25}=sqrt{25(-1)}=sqrt{25i^2}=5i
$$

a. $2i$

b. $-6$

c. 20

d. $5i$

a. No, the graph does not cross the $x$-axis.

b. No, because the corresponding graph does not cross the $x$-axis.

c.
$$
x=dfrac{4pm sqrt{(-4)^2-4(1)(5)}}{2(1)}=dfrac{4pm sqrt{-4}}{2}=dfrac{4pm 2i}{2}=2pm i
$$

d. See (c)

e. Replace $x$ with $2+i$:

$$
y=(2+i)^2-4(2+i)+5=4+4i+i^2-8-4i+5=1+i^2=1-1=0
$$

a. No

b. No

c. $x=2pm i$

d. $x=2pm i$

e. $y=0$

a. The roots are $x=-5$ and $x=-1$.

b. The (double) root is $x=-3$.

c. The function has no roots, because the function does not cross the $x$-axis.

d. Let $h(x)$ be 0:

$$
0=(x+3)^2+4
$$

Subtract 4 from both sides of the equation:

$$
-4=(x+3)^2
$$

Take the square root of both sides of the equation:

$$
pm 2i=pm sqrt{-4}=x+3
$$

Subtract 3 from both sides of the equation:

$$
-3pm 2i=x
$$

e. If the graph intersects the $x$-axis twice, then the function has two real roots. If the graph intersects the $x$-axis once, then the function has one real root. If the graph does not intersect the $x$-axis, then the function has two complex roots.

a. $x=-5$ and $x=1$

b. $x=-3$

c. No

d. $x=-3pm 2i$

e. If the graph intersects the $x$-axis twice, then the function has two real roots. If the graph intersects the $x$-axis once, then the function has one real root. If the graph does not intersect the $x$-axis, then the function has two complex roots.

a. Use distributive property:

$$
2(3+i)=(2)(3)+(2)(i)=6+2i
$$

b. Combine like terms:

$$
10-9i-i=10+(-9-1)i=10-10i
$$

c. Combine like terms:

$$
(-2-7i)+(-1+4i)=(-2-1)+(-7i+4i)=-3-3i
$$

d. Combine like terms:

$$
(-8+2i)-(1-6i)=(-8-1)+(2i-6i)=-9-4i
$$

e. Use $i^2=1$:

$$
(1-i)(8-6i)=8-8i-6i+6i^2=8-6-14i=2-14i
$$

a. $6+2i$

b. $10-10i$

c. $-3-3i$

d. $-9-4i$

e. $2-14i$

a. If you multiply two polynomials, you will always obtain another polynomials and thus the one-variable polynomials are closed under multiplication.

b.
$$
16x^2div 4x=dfrac{16x^2}{4x}=dfrac{16}{4}x^{2-1}=4x
$$

$$
4xdiv 16x^2=dfrac{4x}{16x^2}=dfrac{1}{4x}
$$

Thus we note that the polynomials are not closed under division, because if you divide $4x$ by $16x^2$ you do not obtain a polynomial.

c. Polynomials are closed under addition, subtraction and multiplication, but are not closed under division. Note that this is the same as for integers.

Use that $i^2=-1$:

a.
$$
sqrt{-49}=sqrt{49(-1)}=sqrt{49i^2}=7i
$$

b.
$$
sqrt{-2}=sqrt{2(-1)}=sqrt{2i^2}=sqrt{2}i
$$

c.
$$
(4i)^2=16i^2=16(-1)=-16
$$

d.
$$
3i+2i-5=(3+2)i-5=5i-5
$$

a. $7i$

b. $sqrt{2}i$

c. $-16$

d. $5i-5$

a. Add 49 to both sides of the equation:

$$
49=x^2+18x+81
$$

Factorize:

$$
49=(x+9)^2
$$

Take the square root of both sides of the equation:

$$
pm 7=x+9
$$

Subtract 9 from both sides of the equation (and interchange les and right sides of the equation):

$$
x=-9pm 7=-2text{ or }-16
$$

b. Add 36 to both sides of the equation:

$$
0=x^2+12x+36
$$

Factorize:

$$
0=(x+6)^2
$$

Take the square root of both sides of the equation:

$$
0=x+6
$$

Subtract 6 from both sides of the equation (and interchange les and right sides of the equation):

$$
x=-6
$$

a. $x=-2$ and $x=-16$

b. $x=-6$

b. Use the addition rule:

$$
P(salami text{ or }mayo)=P(salami)+P(mayo)+P(salami text{ and }mayo)
$$

$$
=dfrac{1}{2}+dfrac{2}{3}-dfrac{1}{2}cdot dfrac{2}{3}=dfrac{3}{6}+dfrac{4}{6}-dfrac{2}{6}=dfrac{5}{6}
$$

c. The sum of an event and its complement is 1(=100%)

$$
P(text{not salami and not mayo})=1-P(text{salami or mayo})=1-dfrac{5}{6}=dfrac{1}{6}
$$

d. See (c)

e. There are two outcomes that have both salami and mayonnaise (white and whole grain).

a. Tree diagram

b. $frac{5}{6}$

c. $frac{1}{6}$

d. $frac{1}{6}$

e. white and whole grain

The quadratic formula is
$$
x=dfrac{-bpm sqrt{b^2-4ac}}{2a}
$$

with $a$ the coefficient of $x^2$, $b$ the coefficient of $x$ and $c$ the constant term.

a. The $x$-intercepts seem to be about $x=0.8$ and $x=4.2$:

$$
x=dfrac{5pm sqrt{(-5)^2-4(1)(3)}}{2(1)}=dfrac{5pm sqrt{13}}{2}approx 4.3text{ or } 0.7
$$

b. The $x$-intercepts seem to be about $x=-3.8$ and $x=1.8$:

$$
x=dfrac{-2pm sqrt{(2)^2-4(1)(-6)}}{2(1)}=dfrac{-2pm 2sqrt{7}}{2}=-1pm sqrt{7}approx 1.6text{ or }-3.6
$$

a. $x=frac{5pm sqrt{13}}{2}$

b. $x=-1pmsqrt{7}$

a. Add the powers of the same variables:

$$
(x^2)(x^2y^3)=x^{2+2}y^3=x^4y^3
$$

b. Subtract powers of the same variables:

$$
dfrac{x^3y^4}{x^2y^3}=x^{3-2}y^{4-3}=x^1y^1=xy
$$

c. Add the powers of the same variables:

$$
(2x^2)(-3x^4)=(2)(-3)x^{2+4}=-6x^6
$$

d.
$$
(2x)^3=2^3x^3=8x^3
$$

a. $x^4y^3$

b. $xy$

c. $-6x^6$

d. $8x^3$