Wize High School Grade 12 Pre-Calculus Textbook > Polynomial Functions

Solving Linear & Quadratic Equations

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Solving Linear Equations (Review)
A linear equation is an equation that can be expressed in the form:
ax+b=0\boxed{ax+b=0}ax+b=0​
where a, b∈R a,~b\in\mathbb{R}~a, b∈R  and 'xxx' is the variable. 

Steps to Solving a Linear Equation
If there are any fractions in the equation, eliminate them by multiply the equation by the lowest common denominator (LCD). 
Eliminate parentheses & combine like terms 
Gather all terms with the variable on one side of the equals sign and all other terms on the other side
Divide by the leading coefficient if it is not 1
Verify answer


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Example

Solve x−23+1=2x7\displaystyle \frac{x-2}{3}+1=\frac{2x}{7}3x−2​+1=72x​:

(x−23+1=2x7)×217(x−2)+21=6x7x−14+21=6x(7x−6x)=(14−21)x=−7\displaystyle \begin{array}{r l}
\Bigg(\displaystyle \frac{x-2}{3}+1&=\displaystyle \frac{2x}{7}\Bigg)\times21\\\\
7(x-2)+21&=6x\\\\
7x-14+21&=6x\\\\
(7x-6x)&=(14-21)\\\\
x&=-7
\end{array}(3x−2​+17(x−2)+217x−14+21(7x−6x)x​=72x​)×21=6x=6x=(14−21)=−7​
Verify:
x−23+1=2x7−7−23+1=2(−7)7−93+1=−2−3+1=−2−2=2✓\displaystyle \begin{array}{c l c}
\displaystyle \frac{x-2}{3}+1&=\displaystyle \frac{2x}{7}\\\\
\displaystyle \frac{-7-2}{3}+1&=\displaystyle \frac{2(-7)}{7}\\\\
-\displaystyle \frac{9}{3}+1&=-2\\\\
-3+1&=-2\\\\
-2&=2&\checkmark\end{array}3x−2​+13−7−2​+1−39​+1−3+1−2​=72x​=72(−7)​=−2=−2=2​✓​

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Example: Solving Linear Equations
Find 'xxx':
2x+2=3x−1\displaystyle\frac{2}{x+2}=\frac{3}{x-1}x+22​=x−13​

(2x+2=3x−1)×(x+2)(x−1)2(x−1)=3(x+2)2x−2=3x+6x=−8\begin{array}{rcl}
\Bigg(\displaystyle\frac{2}{x+2}&=&\displaystyle\frac{3}{x-1}\Bigg)\times(x+2)(x-1)\\\\
2(x-1)&=&3(x+2)\\\\
2x-2&=&3x+6\\\\x&=&-8
\end{array}(x+22​2(x−1)2x−2x​====​x−13​)×(x+2)(x−1)3(x+2)3x+6−8​
Verfiy:

2x+2=3x−12−8+2=3−8−1−13=−13✓\begin{array}{rclc}
\displaystyle\frac{2}{x+2}&=&\displaystyle\frac{3}{x-1}\\\\
\displaystyle\frac{2}{-8+2}&=&\displaystyle\frac{3}{-8-1}\\\\
\displaystyle-\frac{1}{3}&=&\displaystyle-\frac{1}{3}&\checkmark
\end{array}x+22​−8+22​−31​​===​x−13​−8−13​−31​​✓​

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Solving Quadratic Equations
A quadratic equation is an equation that can be expressed in the form:
ax2+bx+c=0ax^2+bx+c=0ax2+bx+c=0
where a,b,c,d∈Ra, b, c, d\in\mathbb{R}a,b,c,d∈R and 'xxx' is the variable. 

Algebraic Methods to Solve a Quadratic Equation
Factor
Complete the Square
Quadratic Formula

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Example
Solve 2x2+x−3=02x^2+x-3=02x2+x−3=0 by factoring, completing the square, & using the quadratic formula

Factoring:
2x2+x−3=0(2x2−2x)+(3x−3)=02x(x−1)+3(x−1)=0(x−1)(2x+3)=0∴ x=1, −32\begin{array}{r c l}
2x^2+x-3&=&0\\\\
(2x^2-2x)+(3x-3)&=&0\\\\
2x(x-1)+3(x-1)&=&0\\\\
(x-1)(2x+3)&=&0\\\\
\therefore~x=1,~-\frac{3}{2}

\end{array}2x2+x−3(2x2−2x)+(3x−3)2x(x−1)+3(x−1)(x−1)(2x+3)∴ x=1, −23​​====​0000​

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Completing the Square:
2x2+x−3=0(2x2+x)−3=02(x2+12x+116)−3−18=02(x+14)2=258(x+14)2=2516x+14=±54x=1,−32\begin{array}{rcl}
2x^2+x-3&=&0\\\\
(2x^2+x)-3&=&0\\\\
2\big(x^2+\frac{1}{2}x+\frac{1}{16}\big)-3-\frac{1}{8}&=&0\\\\2\big(x+\frac{1}{4}\big)^2&=&\frac{25}{8}\\\\
\big(x+\frac{1}{4}\big)^2&=&\frac{25}{16}\\\\
x+\frac{1}{4}&=&\pm\frac{5}{4}\\\\x&=&1,-\frac{3}{2}
\end{array}2x2+x−3(2x2+x)−32(x2+21​x+161​)−3−81​2(x+41​)2(x+41​)2x+41​x​=======​000825​1625​±45​1,−23​​

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 Quadratic Formula:
x=−b±b2−4ac2ax=−1±12−4(2)(−3)2(2)x=−1±254x=−1±54x=1, −32\begin{array}{rcl}
x&=&\displaystyle\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\
x&=&\displaystyle\frac{-1\pm\sqrt{1^2-4(2)(-3)}}{2(2)}\\\\
x&=&\displaystyle\frac{-1\pm\sqrt{25}}{4}\\\\
x&=&\displaystyle\frac{-1\pm5}{4}\\\\
x&=&1,~-\frac{3}{2}

\end{array}xxxxx​=====​2a−b±b2−4ac​​2(2)−1±12−4(2)(−3)​​4−1±25​​4−1±5​1, −23​​

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Example: Solving Quadratic Equations (Review)
Solve for 'xxx': −2(x+1)2=−8-2(x+1)^2=-8−2(x+1)2=−8

−2(x+1)2=−8(x+1)2=4(x+1)2=4x+1=±2x=1, −3\begin{array}{rcl}
-2(x+1)^2&=&-8\\\\(x+1)^2&=&4\\\\
\sqrt{(x+1)^2}&=&\sqrt{4}\\\\
x+1&=&\pm2\\\\x&=&1,~-3

\end{array}−2(x+1)2(x+1)2(x+1)2​x+1x​=====​−844​±21, −3​

Find the roots of 3x2−4x−11=03x^2-4x-11=03x2−4x−11=0.

A) 4±2373\displaystyle\frac{4\pm2\sqrt{37}}{3}34±237​​

B)2±2373\displaystyle\frac{2\pm2\sqrt{37}}{3}32±237​​ 

C) 2±373\displaystyle\frac{2\pm\sqrt{37}}{3}32±37​​

D) Not factorable

I don't know

Solve: 
3x+5=(x−1)2−53x+5=(x-1)^2-53x+5=(x−1)2−5

 5±612\displaystyle\frac{5\pm\sqrt{61}}{2}25±61​​ 

3±5−6\displaystyle\frac{3\pm\sqrt{5}}{-6}−63±5​​

 −5±592\displaystyle\frac{-5\pm\sqrt{59}}22−5±59​​ 

−33 and−5-33\text{ and} -5−33 and−5

No Solution

I don't know

Extra Practice

Solving Linear & Quadratic Equations (Review)

Practice: Solving Linear & Quadratic Equations (Review)
Solve: 
(x−3)2+5=2x+4(x-3)^2+5=2x+4(x−3)2+5=2x+4

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)
Find the roots of 5x2−6x−3=05x^2-6x-3=05x2−6x−3=0.

Solving Linear & Quadratic Equations (Review)

Practice: Solving Linear & Quadratic Equations (Review)
Find the roots of 2x2−7x−8=02x^2-7x-8=02x2−7x−8=0.

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