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Solving Quadratic Equations: Examples (page 6 of 6)

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


When you're solving quadratics in your homework, you can often get a "hint" as to the "best" method to use, based on the topic and title of the section. For instance, if you're working on the homework in the "Solving by Factoring" section, then you know that you're supposed to solve by factoring. But in the chapter review and on the test, you don't know which section the quadratic came from. Which method should you use?

You could use the Quadratic Formula for everything, but the Formula can be "overkill". For example:

  • Solve (x + 1)(x – 3) = 0.

    This is a quadratic, and I'm supposed to solve it. I could multiply the left-hand side, simplify to find the coefficients, plug them into the Quadratic Formula, and chug away to the answer.

    But why would I? I mean, for heaven's sake, this is factorable, and they've already factored it and set it equal to zero for me. While the Quadratic Formula would give me the correct answer, why bother with it? Instead, I'll just solve the factors:

      (x + 1)(x – 3) = 0
      x + 1 = 0  or  x – 3 = 0
      x = –1  or  x = 3

      The solution is  x = –1, 3

  • Solve x2 + x – 4 = 0.

    This one doesn't factor (since there are no factors of (1)(–4) = –4 that add to +1), and this isn't formatted as "(squared part) equals (a number)", so I can't use square-rooting to solve. This leaves me with completing the square (yuck!) or the Quadratic Formula. I'll use the Formula:

      x = [-1 ± sqrt(17)] / 2

       

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      The solution is   x = [-1 ± sqrt(17)] / 2

  • Solve x2 – 3x – 4 = 0.

    This one factors easily:

      x2 – 3x – 4 = 0
      (x + 1)(x – 4) = 0
      x + 1 = 0  or  x – 4 = 0
      x = –1  or  x = 4

      The solution is x = –1, 4

  • Solve x2 – 4 = 0.

    This quadratic has just two terms, and nothing factors out of both, so it's a difference of squares (so I can factor) or it can be reformatted as "(squared part) equals (a number)" so I can square-root both sides. In this case, I can factor:

      x2 – 4 = 0
      (x + 2)(x – 2) = 0
      x + 2 = 0  or  x – 2 = 0
      x = –2  or  x = 2

      The solution is x = ± 2

I could have moved the 4 over to the right-hand side of the equation, and then taken the square root of either side of x2 = 4. This method would have given me the exact same answer as the factoring done above. Use whichever method you prefer.

  • Solve 6x2 + 11x – 35 = 0.

    This may factor, but it looks like it would be a fair amount of work, and I'm feeling a bit mindless and lazy at the moment, so I'll use the Quadratic Formula instead:

      x = -7/2, 5/3

    The solutions are fractions, which means the quadratic could have been factored. But I've got my answer, so I don't care any more.

      The solution is x = –7/2, 5/3

  • Solve x2 – 48 = 0.

    This is two terms, and nothing factors out, so either it's a difference of squares (so I can factor) or else it can be formatted as "(squared part equals) a (number)" so I can square-root both sides. Since 48 is not a square, I have to square-root both sides:

      x = ± 4sqrt(3)

      The solution is  x = ± 4sqrt(3)

  • Solve x2 – 7x = 0.

    This quadratic factors easily: Copyright © Elizabeth Stapel 2002-2011 All Rights Reserved

      x2 – 7x = 0
      x(x – 7) = 0
      x = 0  or  x – 7 = 0
      x = 0  or  x = 7

      The solution is x = 0, 7

In general, you first check to see if there is an obvious factoring or if there is an obvious square-rooting that you can do. If not, then it's usually best to resort to the Quadratic Formula. But (warning!) don't only use the Quadratic Formula; while it will always give you the answer — eventually — it is not always the fastest method. And speed can count for a lot on timed tests.

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Cite this article as:

Stapel, Elizabeth. "Solving Quadratic Equations: Examples." Purplemath. Available from
    http://www.purplemath.com/modules/solvquad6.htm. Accessed
 

 

 

 

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