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Solving Quadratic Equations:
  Solving by Completing the Square
(page 3 of 6)

Sections: Solving by: factoring, taking roots, completing the square, using the formula, graphing


The quadratic in the previous section's last example, "(x – 2)2  – 12", can be multiplied out and simplified to be "x2 – 4x – 8". But we would not have been able to solve the equation with the quadratic formatted this way because it doesn't factor and it isn't ready for square-rooting. The only reason we could solve it before was because they'd already put all the x stuff inside a square, so we could square-root both sides. So how do you go from a regular quadratic like "x2 – 4x – 8" to one that is ready to be square-rooted? We would have to "complete the square".

I have a lesson on solving quadratics by completing the square, which explains the steps and gives examples of this process. It also shows how the Quadratic Formula is generated by this process. So I'll just do just one example of the process in this lesson. If you need further instruction, read the lesson at the above hyperlink.

  • Use completing the square to solve x2 – 4x – 8 = 0.

    As noted above, this quadratic does not factor, so I can't solve the equation by factoring. And they haven't given me the quadratic in a form that is ready to square-root. But there is a way for me to manipulate the quadratic to put it into that form, and then solve. It works like this:

    First, I put the loose number on the other side of the equation:

      x2 – 4x – 8 = 0
      x2 – 4x = 8 Copyright © Elizabeth Stapel 2002-2011 All Rights Reserved

    Then I look at the coefficient of the x-term, which is –4 in this case. I take half of this number (including the sign), giving me –2. Then I square this value to get +4, and add this squared value to both sides of the equation:

      x2 – 4x + 4 = 8 + 4
      x2 – 4x + 4 = 12

    This process creates a quadratic that is a perfect square, and factoring gives me:

      (x – 2)2 = 12

       

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    (I know it's a "minus two" inside the parentheses because half of –4 is –2. If you note the sign when you're finding one-half of the coefficient, then you won't mess up the sign when you're converting to squared-binomial form.)

    Now I can square-root both sides of the equation, simplify, and solve:

      (x – 2)2 = 12

      x = 2 ± 2sqrt(3)

    Then the solution is  x = 2 ± 2sqrt(3)

Unless you're told that you have to use completing the square, you will probably not use this method, in actual practice, when solving quadratic equations. Either some other method (such as factoring) will be obvious and quicker, or else the Quadratic Formula (coming up next) will be easier to use. However, if your class covered completing the square, you should expect to be required to show that you can complete the square to solve a quadratic on the next test.

That said, you can probably safely forget this process for solving quadratics after the test, because, instead of completing the square to find messy solutions like the last one above, you'll use a formula....

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Cite this article as:

Stapel, Elizabeth. "Solving Quadratic Equations: Solving by Completing the Square." Purplemath. Available from
    http://www.purplemath.com/modules/solvquad3.htm. Accessed
 

 

 

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