Wize High School Grade 10 Math Textbook > Solving Quadratic Equations

Solving Quadratic Equations by Factoring

ax^2+bx+c

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Solving Quadratic Equations by Factoring
A quadratic equation 0=ax2+bx+c0=ax^2+bx+c0=ax2+bx+c can have 1, 2 or 0 solutions.

How to solve by factoring?
Factor the quadratic expression into 0=(...x+...)(...x+...)0=(...x+...)(...x+...)0=(...x+...)(...x+...)
Let each of these brackets = 0 and solve for xxx

*If you need a refresher or extra practice on factoring, please see the chapter titled "Factoring Polynomials"

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Example: Solving Quadratic Equations by Factoring
Determine the solutions (if any) to the equation 0=x2−2x−350=x^2-2x-350=x2−2x−35

1. We already have 0 on one side of the equation, so we now have to factor the right side of the equation:

2. From the equation, we see that a=1a=1a=1, b=−2b=-2b=−2 and c=−35c=-35c=−35. 
We need two numbers mmm and nnn that multiply to −35-35−35 and add to −2-2−2. 
These numbers are −7-7−7 and 555 
So, the factored form is x2−2x−35=(x−7)(x+5)x^2-2x-35=(x-7)(x+5)x2−2x−35=(x−7)(x+5)

3. Solve the equation:
0=x2−2x−350=(x−7)(x+5)x−7=0orx+5=0x=7x=−5\begin{array}{rcl}
0&=&x^2-2x-35\\
0&=&(x-7)(x+5)\\\\

x-7=0&\text{or}&x+5=0\\
x=7&&x=-5
\end{array}00x−7=0x=7​==or​x2−2x−35(x−7)(x+5)x+5=0x=−5​

4. The solutions are x=7x=7x=7 and x=−5x=-5x=−5
We can check this by substituting each x value into the original equation:
0=(7)2−2(7)−350=49−14−350=0  ✓0=(−5)2−2(−5)−350=25+10−350=0  ✓\begin{array}{ccc}
\begin{array}{rcl}
0&=&(\colorbox{yellow}{$7$})^2-2(\colorbox{yellow}{$7$})-35\\
0&=&49-14-35\\
0&=&0~~\checkmark
\end{array}
&
&
&
\begin{array}{rcl}
0&=&(\colorbox{yellow}{$-5$})^2-2(\colorbox{yellow}{$-5$})-35\\
0&=&25+10-35\\
0&=&0~~\checkmark
\end{array}
\end{array}000​===​(7​)2−2(7​)−3549−14−350  ✓​​​​000​===​(−5​)2−2(−5​)−3525+10−350  ✓​​

Practice: Solving Factored Quadratic Equations
Solve the following equations (find the xxx value that makes the left side of the equation = the right side of the equation)

a) (x−1)(x+4)=0(x-1)(x+4)=0(x−1)(x+4)=0

b) x(x+3)=0x(x+3)=0x(x+3)=0

c) (x−5)2=0(x-5)^2=0(x−5)2=0

a) If there is more than 1 solution for

x

, enter your answer in the form

smaller~value,larger~value

b) If there is more than 1 solution for

x

, enter your answer in the form

smaller~value,larger~value

c) If there is more than 1 solution for

x

, enter your answer in the form

smaller~value,larger~value

I don't know

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Example: Solving Quadratic Equations by Factoring
Determine the solutions (if any) to the equation 3=x2−63=x^2-63=x2−6

1. Rearrange the equation: 0=x2−90=x^2-90=x2−9

2. From the equation, we see that this is a difference of squares
Rewriting the quadratic expression on the right: x2−32x^2-3^2x2−32
Recall the difference of squares formula a2−b2=(a−b)(a+b)a^2-b^2=(a-b)(a+b)a2−b2=(a−b)(a+b)
The factored form is (x−3)(x+3)(x-3)(x+3)(x−3)(x+3)

3. Solve the equation:
0=x2−90=(x−3)(x+3)x−3=0orx+3=0x=3x=−3\begin{array}{rcl}
0&=&x^2-9\\
0&=&(x-3)(x+3)\\\\

x-3=0&\text{or}&x+3=0\\
x=3&&x=-3
\end{array}00x−3=0x=3​==or​x2−9(x−3)(x+3)x+3=0x=−3​

4. The solutions are x=3x=3x=3 and x=−3x=-3x=−3
We can check this by substituting each x value into the original equation:
0=(3)2−90=9−90=0  ✓0=(−3)2−90=9−90=0  ✓\begin{array}{ccc}
\begin{array}{rcl}
0&=&(\colorbox{yellow}{$3$})^2-9\\
0&=&9-9\\
0&=&0~~\checkmark
\end{array}
&
&
&
\begin{array}{rcl}
0&=&(\colorbox{yellow}{$-3$})^2-9\\
0&=&9-9\\
0&=&0~~\checkmark
\end{array}
\end{array}000​===​(3​)2−99−90  ✓​​​​000​===​(−3​)2−99−90  ✓​​

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Example: Solving Quadratic Equations by Factoring
Determine the solutions (if any) to the equation 2=4x2+4x+32=4x^2+4x+32=4x2+4x+3

1. Rearrange the equation: 0=4x2+4x+10=4x^2+4x+10=4x2+4x+1

2. From the equation, we see that a=4a=4a=4, b=4b=4b=4 and c=1c=1c=1
We need two numbers mmm and nnn that multiply to 4×1=44\times1=44×1=4 and add to 444
These numbers are 222 and 222
Rewrite the quadratic expression: 4x2+2x+2x+14x^2+2x+2x+14x2+2x+2x+1
Factor this expression: (4x2+2x)+(2x+1)=2x(2x+1)‾+1(2x+1)‾=(2x+1)(2x+1)=(2x+1)2\begin{array}{rl}
&(4x^2+2x)+(2x+1)\\
=&2x\underline{(2x+1)}+1\underline{(2x+1)}\\
=&(2x+1)(2x+1)\\
=&(2x+1)^2
\end{array}===​(4x2+2x)+(2x+1)2x(2x+1)​+1(2x+1)​(2x+1)(2x+1)(2x+1)2​
*Alternative, you could have recognized this as a perfect square trinomial and factored it using the short-cut: a2+2ab+b2=(a+b)2a^2+2ab+b^2=(a+b)^2a2+2ab+b2=(a+b)2

3. Solve the equation:
0=(2x+1)20=(2x+1)(2x−1)0=2x+1−1=2x−12=x\begin{array}{rcl}
0&=&(2x+1)^2\\
0&=&(2x+1)(2x-1)\\

0&=&2x+1\\
-1&=&2x\\
-\dfrac{1}{2}&=&x
\end{array}000−1−21​​=====​(2x+1)2(2x+1)(2x−1)2x+12xx​

4. There is only one solution x=−12x=-\dfrac{1}{2}x=−21​
We can check this by substituting each x value into the original equation:
2=4(−12)2+4(−12)+32=4(14)+4(−12)+32=1−2+32=2  ✓\begin{array}{ccc}
\begin{array}{rcl}
2&=&4\left(\colorbox{yellow}{$-\dfrac{1}{2}$}\right)^2+4\left(\colorbox{yellow}{$-\dfrac{1}{2}$}\right)+3\\
2&=&4\left(\dfrac{1}{4}\right)+4\left(-\dfrac{1}{2}\right)+3\\
2&=&1-2+3\\
2&=&2~~\checkmark
\end{array}

\end{array}2222​====​4(−21​​)2+4(−21​​)+34(41​)+4(−21​)+31−2+32  ✓​​

Practice: Solving Quadratic Equations by Factoring
Determine the roots (zeros or x-intercepts) of the following equations:
a) 4=x2−4x−14=x^2-4x-14=x2−4x−1

b) 7=6x2−11x−37=6x^2-11x-37=6x2−11x−3

a) If there is more than 1 root, enter your answer in the form

smaller~value,~larger~value

b) If there is more than 1 root, enter your answer in the form

smaller~value,~larger~value

I don't know

Practice: Solving Quadratic Equations
A local movie theatre's profit is modeled by the equation P=−10x2+220x−210P=-10x^2+220x-210P=−10x2+220x−210, where PPP represents the profit on any given day, and xxx represents the ticket prices.

a) At what ticket price(s) will the movie theatre break even? (the theatre does not make money and does not lose money)

b) Determine the movie ticket price that maximizes profit, and determine the maximum profit.

c) At what ticket price(s) will the profit be $640?

a) What is the lowest ticket price that will break even?

a) What is the highest ticket price that will break even?

b) What is the maximum profit?

b) What is the ticket price that will maximize profit?

c) What is the LOWEST ticket price that produces a profit of $640?

c) What is the HIGHEST ticket price that produces a profit of $640?

I don't know

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Practice: Applications with Quadratic Equations
Find the intersection between the line y=3x−6y=3x-6y=3x−6 and the parabola y=6x2+31x−16y=6x^2+31x-16y=6x2+31x−16, algebraically.

I have answered this question

Extra Practice

Solving Linear & Quadratic Equations (Review)

Practice: Solving Linear & Quadratic Equations (Review)
Solve: 
(x+2)2−3=x+1(x+2)^2-3=x+1(x+2)2−3=x+1

Solving Linear & Quadratic Equations (Review)

Practice: Solving Linear & Quadratic Equations (Review)
Solve for 'x':
x2−4x−21=0x^2-4x-21=0x2−4x−21=0

Solving Linear & Quadratic Equations (Review)

Practice: Solving Linear & Quadratic Equations (Review)
Solve for 'x':
x2+2x−99=0x^2+2x-99=0x2+2x−99=0

Wize High School Grade 10 Math Textbook > Solving Quadratic Equations

Solving Quadratic Equations by Factoring

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