All Solutions
Section 3-5: Quadratic Function Models: Solving Quadratic Equations
$ax^2+bx+c=0$
For $a=1$, then we just need find the factors of $c$ whose sum is $b$.
$c=scdot q$
$b=s+q$
The factors would be
$(x+s)(x+q)$
If $aneq 1$
Find the factors of $a$ and $c$ such that
$a=pcdot r$
$c=qcdot s$
$b=pcdot s+rcdot q$
Then, the factors would be
$(px+q)(rx+s)$
$x^2+5x+4=(x+1)(x+4)$
$x+1=0implies x=-1$
$x+4=0implies x=-4$
Therefore, $x=-1 text{ or }-4$
$x^2-11x+18=(x-9)(x-2)$
$x=9$ or $x=2$
Use difference of two squares $a^2-b^2=(a+b)(a-b)$
$[(2x)^2-3^2]=(2x+3)(2x-3)$
$2x+3=0implies x=-dfrac{3}{2}$
$2x-3=0implies x=dfrac{3}{2}$
Therefore, $x=-dfrac{3}{2};text{or};dfrac{3}{2}$
$2x+1=0implies x=-dfrac{1}{2}$
$x-4=0implies x=4$
Therefore, $x=-dfrac{1}{2}$ or $4$
b) $x=2$ or 9
c) $x=pmdfrac{3}{2}$
d) $x=-dfrac{1}{2}$ or 4
$$
x=dfrac{-bpmsqrt{b^2-4ac}}{2a}
$$
{text{a) }}a = 1{text{ , }}b = – 4,{text{ }}c = – 9 hfill \
hfill \
x = frac{{ – left( { – 4} right) pm sqrt {{{left( { – 4} right)}^2} – 4left( 1 right)left( { – 9} right)} }}{{2left( 1 right)}} hfill \
hfill \
x = frac{{4 pm sqrt {16 + 36} }}{2} hfill \
x = frac{{4 pm sqrt {52} }}{2} hfill \
x = frac{{4 pm sqrt {4 times 13} }}{2} hfill \
x = frac{{4 pm 2sqrt {13} }}{2} hfill \
x = 2 pm sqrt {13} hfill \
x=-5.61 text{ or } -1.61 hfill \
end{gathered} ]
{text{b) }}a = 3{text{ , }}b = 2,{text{ }}c = – 8 hfill \
hfill \
x = frac{{ – 2 pm sqrt {{{left( 2 right)}^2} – 4left( 3 right)left( { – 8} right)} }}{{2left( 3 right)}} hfill \
hfill \
x = frac{{-2 pm sqrt {4 + 96} }}{6} hfill \
x = frac{{-2 pm sqrt {100} }}{6} hfill \
x = frac{{-2 pm 10}}{6} hfill \
x = -2 {text{ or }}1.33 hfill \
end{gathered} ]
{text{c)}},, – 2{x^2} + 3x – 6 = 0 hfill \
a = – 2,{text{ }}b = 3,{text{ }}c = – 6 hfill \
hfill \
x = frac{{ – 3 pm sqrt {{3^2} – 4left( { – 2} right)left( { – 6} right)} }}{{2left( { – 2} right)}} hfill \
x = frac{{ – 3 pm sqrt {9 – 48} }}{{ – 4}} hfill \
x = frac{{ – 3 pm sqrt { – 39} }}{{ – 4}} hfill \
hfill \
{text{Since the expression inside the radical sign is }} hfill \
{text{negative, there are no real roots}}{text{.}} hfill \
end{gathered} ]
{text{d)}},, – 2{x^2} + 3x – 6 = 0 hfill \
a = 0.5,{text{ }}b = – 2.2{text{ }}c = – 4.7 hfill \
hfill \
x = frac{{ – left( { – 2.2} right) pm sqrt {{{left( {2.2} right)}^2} – 4left( {0.5} right)left( { – 4.7} right)} }}{{2left( {0.5} right)}} hfill \
x = frac{{2.2 pm sqrt {frac{{356}}{{25}}} }}{1} hfill \
x = 2.2 pm sqrt {frac{{4 times 89}}{{25}}} hfill \
x = 2.2 pm sqrt {frac{{{2^2} times 89}}{{{5^2}}}} hfill \
x = 2.2 pm frac{2}{5}sqrt {89} hfill \
x = – 1.57{text{ or 5}}{text{.97}} hfill \
end{gathered} ]
b) $x=1.33$ or $-2$
c) no real roots
d) $x=-1.57$ or $5.97$
{text{a)}} hfill \
2{x^2} – 3x = {x^2} + 7x hfill \
{x^2} – 10x = 0 hfill \
xleft( {x – 10} right) = 0 hfill \
x = 0{text{ or 10}} hfill \
end{gathered} ]
{text{b)}} hfill \
4{x^2} + 6x + 1 = 0 hfill \
a = 4,{text{ }}b = 6,{text{ }}c = 1 hfill \
x = frac{{ – 6 pm sqrt {{6^2} – 4left( 4 right)left( 1 right)} }}{{2left( 4 right)}} hfill \
x = frac{{ – 6 pm sqrt {36 – 16} }}{8} hfill \
x = frac{{ – 6 pm sqrt {20} }}{8} hfill \
x = frac{{ – 6 pm sqrt {4 times 5} }}{8} hfill \
x = frac{{ – 6 pm 2sqrt 5 }}{8} hfill \
x = frac{{ – 3 pm sqrt 5 }}{4} hfill \
end{gathered} ]
$$
x=dfrac{-bpmsqrt{b^2-4ac}}{2a}
$$
{text{c) }}{x^2} + 4x – 3 = 0 hfill \
a = 1,{text{ }}b = 4,{text{ }}c = – 3 hfill \
hfill \
x = frac{{ – b pm sqrt {{b^2} – 4ac} }}{{2a}} hfill \
x = frac{{ – 4 pm sqrt {{4^2} – 4left( 1 right)left( { – 3} right)} }}{{2left( 1 right)}} hfill \
x = frac{{ – 4 pm sqrt {28} }}{2} hfill \
x = frac{{ – 4 pm sqrt {4 times 7} }}{2} hfill \
x = frac{{ – 4 pm 2sqrt 7 }}{2} hfill \
x = – 2 pm sqrt 7 hfill \
end{gathered} ]
$$
x=dfrac{-bpmsqrt{b^2-4ac}}{2a}
$$
{text{d)}} hfill \
{left( {x + 3} right)^2} = – 2x hfill \
{x^2} + 6x + 9 = – 2x hfill \
{x^2} + 8x + 9 = 0 hfill \
a = 1,{text{ }}b = 8,{text{ }}c = 9 hfill \
hfill \
x = frac{{ – 8 pm sqrt {{8^2} – 4left( 1 right)left( 9 right)} }}{{2left( 1 right)}} hfill \
x = frac{{ – 8 pm sqrt {28} }}{2} hfill \
x = frac{{ – 8 pm sqrt {4 times 7} }}{2} hfill \
x = frac{{ – 8 pm 2sqrt 7 }}{2} hfill \
x = – 4 pm sqrt 7 hfill \
end{gathered} ]
$$
x=dfrac{-bpmsqrt{b^2-4ac}}{2a}
$$
{text{e)}} hfill \
3{x^2} – 5x = 2{x^2} + 4x + 10 hfill \
{x^2} – 9x – 10 = 0 hfill \
left( {x – 10} right)left( {x + 1} right) = 0 hfill \
x = 10{text{ or }} – 1 hfill \
end{gathered} ]
{text{f)}} hfill \
{text{2}}left( {x + 3} right)left( {x – 4} right) = 6x + 6 hfill \
2left( {{x^2} – x – 12} right) = 6x + 6 hfill \
2{x^2} – 2x – 24 = 6x + 6 hfill \
2{x^2} – 8x – 30 = 0 hfill \
a = 2,{text{ }}b = – 8,{text{ }}c = – 30 hfill \
x = frac{{ – left( { – 8} right) pm sqrt {{{left( { – 8} right)}^2} – 4left( 2 right)left( { – 30} right)} }}{{2left( 2 right)}} hfill \
x = frac{{8 pm sqrt {304} }}{4} hfill \
x = frac{{8 pm sqrt {16 times 19} }}{4} hfill \
x = frac{{8 pm 4sqrt {19} }}{4} hfill \
x = 2 pm sqrt {19} hfill \
end{gathered} ]
$$
x=dfrac{-bpmsqrt{b^2-4ac}}{2a}
$$
b) quadratic formula ; $x=dfrac{-3pmsqrt{5}}{4}$
c) quadratic formula ; $x=-2pmsqrt{7}$
d) quadratic formula ; $x=-4pmsqrt{7}$
e) factoring ; $x=-1$ or 10
f) quadratic formula ; $x=2pmsqrt{19}$
{text{The }}x{text{-intercepts are the point on the graph }} hfill \
{text{that touches the }}x{text{-axis, which corresponds}} hfill \
{text{to the values of }}x{text{ when }}fleft( x right) = 0 hfill \
{text{a) We can use quadratic formula to find these}} hfill \
{text{values of }}x. hfill \
hfill \
3{x^2} – 7x – 2 = 0 hfill \
a = 3,{text{ }}b = – 7,{text{ }}c = – 2 hfill \
x = frac{{ – b pm sqrt {{b^2} – 4ac} }}{{2a}} hfill \
x = frac{{ – left( { – 7} right) pm sqrt {{{left( { – 7} right)}^2} – 4left( 3 right)left( { – 2} right)} }}{{2left( 3 right)}} hfill \
x = frac{{7 pm sqrt {49 + 24} }}{6} hfill \
x = frac{{7 pm sqrt {73} }}{6} hfill \
x = 2.59{text{ or }} – {text{0}}{text{.26}} hfill \
hfill \
{text{The }}x{text{-intercepts are }}left( {2.59,0} right){text{ and }}left( { – 0.26,0} right) hfill \
end{gathered} ]
{text{b) We can use factoring to find these}} hfill \
{text{values of }}x. hfill \
hfill \
– 4{x^2} + 25x – 21 = 0 hfill \
4{x^2} – 25x + 21 = 0 hfill \
left( {4x – 21} right)left( {x – 1} right) hfill \
x = frac{{21}}{4}{text{ or 1}} hfill \
hfill \
{text{The }}x{text{-intercepts are }}left( {frac{{21}}{4},0} right){text{ and }}left( {1,0} right) hfill \
end{gathered} ]
b) $(1,0), left(dfrac{21}{4},0right)$
To find such values of $x$, we can use quadratic formula or factoring if applicable.
a) $P(x)=-x^2+12x+28$
$P(x)=-(x+2)(x-14)$
$x=-2$ or 14
Note that $x$ is in units of thousands.
Since sold quantities can’t be negative, the break-even quantity is $14,000$
$P(x)=-2(x^2+9x-20)$
$P(x)=-2(x-5)(x-4)$
$x=5$ or 4
The break-even quantity is $5000$ or $4000$
$a=-2$ , $b=22$ , $c=-17$
This cannot be factored so we shall use quadratic formula
$x=dfrac{-bpmsqrt{b^2-4ac}}{2a}$
$x=dfrac{-22pmsqrt{22^2-4(-2)(-17)}}{2(-2)}$
$x=0.836$ or 10.164
Thus, the break-even quantity is 836 or 10,164.
$a=-0.5$ , $b=6$ , $c=-5$
This cannot be factored so we shall use quadratic formula.
$x=dfrac{-bpmsqrt{b^2-4ac}}{2a}$
$x=dfrac{-6pmsqrt{6^2-4(-0.5)(-5)}}{2(-0.5)}$
$x=11.099$ or 0.9010
Thus, the break-even quantity is $11,099$ or $901$
b) $4000$ or $5000$
c) $836$ or $10,164$
d) $901$ or $11,099$
To find the time at which the ball hits the ground, we shall find the value of $t$ when $h(t)=0$
$h(t)=-4.9t^2+6t+0.6$
This is not factorable so we shall use quadratic formula.
$a=-4.9$ , $b=6$ , $c=0.6$
$t=dfrac{-bpmsqrt{b^2-4ac}}{2a}=dfrac{6pmsqrt{6^2-4(-4.9)(0.6)}}{2(-4.9)}$
$t=-0.09$ or 1.32
In this case, time cannot take negative values,
therefore, the time at which the ball hits the ground is $1.32$ s
$P(t)=0.4t^2+10t+50$
where $t$ is time in years since $1995$
and $P(t)$ is population in thousands.
$P(0)=0.4(0)^2+10(0)+50$
$P(0)=50$
The population in 1995 is $50,000$
$P(15)=0.4(15)^2+10(15)+50$
$P(15)=290$
The population in 2010 will be $290,000$
$450=0.4t^2+10t+50$
$0.4t^2+10t-400=0$
$a=0.4$ , $b=10$, $c=-400$
Use quadratic formula
$x=dfrac{-bpmsqrt{b^2-4ac}}{2a}$
$x=dfrac{-10pmsqrt{10^2-4(0.4)(-400)}}{2(0.4)}$
$x=-46.5$ or $21.5$
$t=1995+21.5=2016.5approx 2017$
Therefore, the population will be at least 450,000 in the year 2017
b) $290,000$
c) 2017
Let $x$ be the width, it follows that the length is $x+7$
A $=$ length $times$ width
$350= (x+7)(x)$
$350=x^2+7x$
$x^2+7x-350=0$
$(x-15)(x+22)=0$
$x=15$ or $-22$
Side lengths can’t be negative, so
width = 15 m
length = 15+7 = 22 m
Therefore, the rectangle has dimensions 15 m by 22 m
It is mentioned that the sum of squares is $685$, thus, we can write this as
$x^2+(x+1)^2=685$
$x^2+x^2+2x+1=685$
$2x^2+2x-684=0$
We can solve this by factoring
$2(x^2+x-342)=0$
$(x-18)(x+19)=0$
$x=18implies x+1=19$
or
$x=-19implies x+1=-18$
The two consecutive numbers are either $-19,-18$ or $18,19$
The area of a triangle with base $b$ and height $h$ is
$A=dfrac{1}{2}bh$
$96=dfrac{1}{2}(x)(2x+8)$
$$
96=x(x+4)
$$
$96=x^2+4x$
$x^2+4x-96=0$
$(x-8)(x+12)=0$
$x=8$ or $-$12
Lengths can’t have negative values so choose $x=8$
base = 8
height = $2x+8 = 2(8)+8=24$
Therefore, the triangle has a base of 8 cm and a height of 24 cm.
After mowing, it would have a length of $25-2x$ and a width of $15-2x$
It is mentioned that the lawn is $60%$ of its original area.
The formula for area of rectangle is length $times$ width
The original area is $25times 15=375$
The new area is
$375(0.6)=(25-2x)(15-2x)$
$225=375-50x-30x+4x^2$
$4x^2-80x+150=0$
$2(2x^2-40x+75)=0$
$2x^2-40x+75=0$
We cannot factor this so we shall use quadratic formula
$x=dfrac{-(-40)pmsqrt{(-40)^2-4(2)(75)}}{2(2)}$
$x=17.91$ or 2.1
If $x=17.91,;15-2x<0$ which is not possible, thus, choose $x=2.1$
$h(t)=-4.9t^2+92t+9$
We shall find $t$ such that $h(t)=150$
$150=-4.9t^2+92t+9$
$4.9t^2-92t+141=0$
We shall use quadratic formula
$a=4.9$ , $b=-92$, $c=141$
$x=dfrac{-bpmsqrt{b^2-4ac}}{2a}$
$x=dfrac{-(-92)pmsqrt{(-92)^2-4(4.9)(141)}}{2(4.9)}$
$x=1.68$ or 17.09
Therefore, it would a height of 150 m at 1.68 s and 17.09 s
$17.09-1.68=15.41$ s
b) $15.41$ s
Fare price = $2+0.15x$
Number of Passengers = $4000-40x$
The target revenue is $$10,450$
$(2+0.15x)(4000-40x)=10,450$
$8000-520x-6x^2=10,450$
$6x^2-540x+2450=0$
$2(3x^2-270x+1225)=0$
Use quadratic formula
$a=3$ , $b=-270$ , $c=1225$
$x=dfrac{-bpmsqrt{b^2-4ac}}{2a}$
$x=dfrac{-(-270)pmsqrt{(-270)^2-4(3)(-1225)}}{2(3)}$
$x=85.21$ or 4.79
price = $2+0.15x=$14.75$
number of passengers = $4000-40x=600$
If $x=5$
price = $2+0.15x=$2.75$
number of passengers = $4000-40x=3800$
Both cases are possible but to cater more passengers, we shall use $$2.75$.
$2.75
$$
$f(x)=-2x^2+14x-24$
$-2(x^2-7x+12)=0$
$-2(x-4)(x-3)=0$
$x=4$ or $x=3$
$a=-2$, $b=14$, $c=-24$
$x=-dfrac{bpmsqrt{b^2-4ac}}{2a}$
$x=dfrac{-14pmsqrt{14^2-4(-2)(-24)}}{2(-2)}$
$x=dfrac{-14pm sqrt{4}}{-4}$
$x=dfrac{-14pm 2}{-4}$
$x=4$ or 3
b) using quadratic formula
c) using graphing calculator
Use the Pythagorean formula $c^2=a^2+b^2$
Also remember that $(a+b)^2=a^2+2ab+b^2$
$(2x+6)^2=x^2+(54-3x)^2$
$(2x)^2+2(2x)(6)+6^2=x^2+54^2-2(54)(3x)+(3x)^2$
$4x^2+24x+36=x^2+2916-324x+9x^2$
$6x^2-348x+2880=0$
$6(x^2-58x+480)=0$
$6(x-48)(x-10)=0$
$x=48$ or 10
If $x=48$ , $54-3x<0$, so choose only $x=10$
$54-3x=24$
$2x+6=26$
Therefore, the triangle has sides $10$ cm , 24 cm , 26 cm
$3x-1+dfrac{1}{x+1}=0$
Multiply both sides by $(x+1)$
$3x(x+1)-(x+1)+1=0$
$3x^2+3x-x-1+1=0$
$3x^2+2x=0$
$x(3x+2)=0$
$x=0$ or $3x+2=0$
$x=0$ or $x=-dfrac{2}{3}$
