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Study Guides > MATH 1314: College Algebra

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.
  1. 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]
  2. 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]
  3. 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]
  4. 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]
  5. 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]
  6. 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

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