# Completing the Square

Not all quadratic equations can be factored or can be solved in their original form using the square root property. In these cases, we may use a method for solving a**quadratic equation**known as

**completing the square**. Using this method, we add or subtract terms to both sides of the equation until we have a perfect square trinomial on one side of the equal sign. We then apply the square root property. To complete the square, the leading coefficient,

*a*, must equal 1. If it does not, then divide the entire equation by

*a*. Then, we can use the following procedures to solve a quadratic equation by completing the square. We will use the example [latex]{x}^{2}+4x+1=0[/latex] to illustrate each step.

- Given a quadratic equation that cannot be factored, and with [latex]a=1[/latex], first add or subtract the constant term to the right sign of the equal sign.
[latex]{x}^{2}+4x=-1[/latex]
- Multiply the
*b*term by [latex]\frac{1}{2}[/latex] and square it.[latex]\begin{array}{l}\frac{1}{2}\left(4\right)=2\hfill \\ {2}^{2}=4\hfill \end{array}[/latex] - Add [latex]{\left(\frac{1}{2}b\right)}^{2}[/latex] to both sides of the equal sign and simplify the right side. We have
[latex]\begin{array}{l}{x}^{2}+4x+4=-1+4\hfill \\ {x}^{2}+4x+4=3\hfill \end{array}[/latex]
- The left side of the equation can now be factored as a perfect square.
[latex]\begin{array}{l}{x}^{2}+4x+4=3\hfill \\ {\left(x+2\right)}^{2}=3\hfill \end{array}[/latex]
- Use the square root property and solve.
[latex]\begin{array}{l}\sqrt{{\left(x+2\right)}^{2}}=\pm \sqrt{3}\hfill \\ x+2=\pm \sqrt{3}\hfill \\ x=-2\pm \sqrt{3}\hfill \end{array}[/latex]
- The solutions are [latex]x=-2+\sqrt{3}[/latex], [latex]x=-2-\sqrt{3}[/latex].

### Example 8: Solving a Quadratic by Completing the Square

Solve the quadratic equation by completing the square: [latex]{x}^{2}-3x - 5=0[/latex].### Solution

First, move the constant term to the right side of the equal sign.[latex]{x}^{2}-3x=5[/latex]

Then, take [latex]\frac{1}{2}[/latex] of the *b*term and square it.

[latex]\begin{array}{l}\frac{1}{2}\left(-3\right)=-\frac{3}{2}\hfill \\ {\left(-\frac{3}{2}\right)}^{2}=\frac{9}{4}\hfill \end{array}[/latex]

Add the result to both sides of the equal sign.
[latex]\begin{array}{l}\text{ }{x}^{2}-3x+{\left(-\frac{3}{2}\right)}^{2}=5+{\left(-\frac{3}{2}\right)}^{2}\hfill \\ {x}^{2}-3x+\frac{9}{4}=5+\frac{9}{4}\hfill \end{array}[/latex]

Factor the left side as a perfect square and simplify the right side.
[latex]{\left(x-\frac{3}{2}\right)}^{2}=\frac{29}{4}[/latex]

Use the square root property and solve.
[latex]\begin{array}{l}\sqrt{{\left(x-\frac{3}{2}\right)}^{2}}\hfill&=\pm \sqrt{\frac{29}{4}}\hfill \\ \left(x-\frac{3}{2}\right)\hfill&=\pm \frac{\sqrt{29}}{2}\hfill \\ x\hfill&=\frac{3}{2}\pm \frac{\sqrt{29}}{2}\hfill \end{array}[/latex]

The solutions are [latex]x=\frac{3}{2}+\frac{\sqrt{29}}{2}[/latex], [latex]x=\frac{3}{2}-\frac{\sqrt{29}}{2}[/latex].
### Try It 7

Solve by completing the square: [latex]{x}^{2}-6x=13[/latex]. Solution## Licenses & Attributions

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