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Wize High School Algebra I Textbook (Common Core) > Solving Quadratic Equations

Solving Quadratic Equations by Completing the Square

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Solving Quadratic Equations by Completing the Square

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

How to solve by completing the square?

  1. Complete the square to turn the expression into vertex form 0=a(x−h)2+k0=a(x-h)^2+k
  2. Solve the equation by using reverse BEDMAS (reverse order of operations)
  3. Interpret your solutions

*If you need a refresher or extra practice on completing the square, please see the chapter titled "Completing the Square".

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Finding the Square Root of x2\bco {x^2}

When we take the square root of x2x^2 , we actually get two different answers!

Example 1
To solve x2=1x^2=1, we find the square root of both side, and get x=1x=1. We can check this answer by putting it back into the equation:
x2=1(1)2=11=1  ✓\begin{array}{rcl} x^2&=&1\\ (\colorbox{yellow}{$1$})^2&=&1\\ 1&=&1~~\checkmark \end{array}
However, notice that x=−1x=-1 is also a solution!
x2=1(−1)2=11=1  ✓\begin{array}{rcl} x^2&=&1\\ (\colorbox{yellow}{$-1$})^2&=&1\\ 1&=&1~~\checkmark \end{array}

We use the ±\pm symbol to represent a "plus or minus" answer.

So, the solution to x2=1x^2=1 is x=±1\boxed{x=\colorbox{yellow}{$\pm$} 1}.

Example 2
Solve x2=25x^2=25.

x2=25x2=25x=±5\begin{array}{rcl} x^2&=&25\\ \sqrt{x^2}&=&\sqrt{25}\\ x&=&\colorbox{yellow}{$\pm$}5 \end{array}
So, xx could be 55 or −5-5.
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Example: Solving Quadratic Equations

Solve the equation 3=(x−6)2−133=(x-6)^2-13.

3=(x−6)2−13+13                 +1316=(x−6)216=(x−6)2±4=x−6\begin{array}{rcl} 3&=&(x-6)^2-13\\ \scriptsize\colorTwo{+13}&&~~~~~~~~~~~~~~~~~\scriptsize\colorTwo{+13}\\[1em] 16&=&(x-6)^2\\[1em] \sqrt{16}&=&\sqrt{(x-6)^2}\\[1em] \colorbox{yellow}{$\pm$}4&=&x-6\\ \end{array}

So, we get two answers from here:
4=x−610=x\begin{array}{rcl} 4&=&x-6\\ 10&=&x \end{array} or −4=x−62=x\begin{array}{rcl} -4&=&x-6\\ 2&=&x \end{array}

Therefore, the solutions (answers) are x=10x=10 or x=2x=2

Let's check our answers:
3=(x−6)2−133=(10−6)2−133=(4)2−133=16−133=3  ✓3=(x−6)2−133=(2−6)2−133=(−4)2−133=16−133=3  ✓\begin{array}{cccc} \begin{array}{rcl} 3&=&(x-6)^2-13\\ 3&=&(\colorbox{yellow}{$10$}-6)^2-13\\ 3&=&(4)^2-13\\ 3&=&16-13\\ 3&=&3~~\checkmark \end{array} &&&& \begin{array}{rcl} 3&=&(x-6)^2-13\\ 3&=&(\colorbox{yellow}{$2$}-6)^2-13\\ 3&=&(-4)^2-13\\ 3&=&16-13\\ 3&=&3~~\checkmark \end{array} \end{array}

Practice: Solving Quadratic Equations

Solve the following quadratic equations.

a) 0=x2−40=x^2-4

b) 0=(n−5)2−160=(n-5)^2-16

c) −2=(t+1)2−23-2=(t+1)^2-23

Practice: Solving Quadratic Equations

Solve 113=2x2−12x+3113=2x^2-12x+3 by first completing the square.

Practice: Solving Quadratic Equations

The height hh (in feet) of a baseball in the air tt seconds after it is hit by the bat is given by the equation h=−3t2+24t−20h=-3t^2+24t-20.

a) What is the maximum height this baseball reaches and when does that occur?

b) When does the ball reach 16 feet?


Practice: Completing the Square

The height of a bungee jumper is given by h=t2−10t+35h=t^2-10t+35. Where hh is the height (in meters) and tt is the time since the bungee jumper leaves the platform (in seconds). When does the bungee jumper first reach 19 meters?