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Solving Radical Equations: Examples (page 2 of 6)


  • Solve the equation:
    • sqrt(x – 1) = x – 7

    The two lines represented by the two sides of this equation are:

      y = sqrt(x – 1); y = x – 7

    ...and they graph as: Copyright Elizabeth Stapel 2002-2011 All Rights Reserved

      graph

    ...so you can see that there should be a solution at or about x = 10. To solve this algebraically, I need to square each side:

      sqrt(x – 1) = x – 7; (sqrt(x – 1))^2 = (x – 7)^2
            x 1 = (x 7)(x 7)
            x 1 = x2 14x + 49

    The squared expressions can be graphed as the lines y = x 1 and y = x2 14x + 49. The solutions of x 1 = x2 14x + 49 are the intersection points of the two lines:

      graph

    As you can see, the intersection point at x = 10, from the first graph, is still there, but now a second, extraneous, solution has appeared at x = 5! ("Extraneous", pronounced as "ek-STRAY-nee-uss", in this context means "mathematically correct, but not relevant or useful, as far as the original question is concerned".) Continuing the solution:

      x 1 = x2 14x + 49
      0 = x2 15x + 50

      0 = (x 5)(x 10)

      x = 5, x = 10

    So I got the result that the second graph led me to expect, but I also know, from the first graph, that "x = 5" should not be a solution. This again illustrates why you always need to check your answers when solving radical equations: the very act of squaring has, in this case, produced an extra and incorrect "solution". Here's my check:

     

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      x = 5:

        2 is not equal to -2

      x = 10:

        3 = 3

    So the answer is x = 10.

  • Solve the equation:
    • sqrt(x - 2) = 5

    Since this equation is in the form "(square root) = (number)", I can proceed directly to squaring both sides:

      (sqrt(x - 2))^2 = 5^2
             x 2 = 25
                  x = 27

    This solution matches what I would expect from the graph of the two sides of the equation:

graph

    As you can see above, the lines:

      y = sqrt(x - 2)
      y = 5

    ...intersect at x = 27, as the algebra had already shown me. Checking, I get:

      5 = 5

    So the solution is x = 27.

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Cite this article as:

Stapel, Elizabeth. "Solving Radical Equations: Examples." Purplemath. Available from
    http://www.purplemath.com/modules/solverad2.htm. Accessed
 

 



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