Return to the Purplemath home page

 

Try a MathHelp.com demo lesson Join MathHelp.com Login to MathHelp.com

 

Index of free lessons | PM's free lessons in offline form | Forums | Search free lessons | Print this page | Local tutors

X

Solving Radical Equations: Examples (page 5 of 6)


  • Solve the equation:  sqrt(2x^2 – 7) = 3 – x
  • Square both sides, being careful to write out the square on the right-hand side:

      (x + 8)(x – 2) = 0

    Then x = –8 and x = 2. Are both of these solutions valid? Graphing the lines for either side of the original equation:

      y = sqrt(2x^2) – 7; y = 3 – x

    ...I get the following graph:

      graph

    It appears that both solutions are valid. Here's the check:

    x = –8:
     

     

    x = 2:
     

      11 = 11

     

      1 = 1

    So the solution is x = –8 or x = 2.


The following examples are not complete. I'll leave the checking to you!

  • Solve:  sqrt(2x + 9) – sqrt(x + 1) = sqrt(x + 4)
  • This equation will have to be squared twice in order to solve it:

      0 = x(x + 5)

It appears that the solutions are x = –5 and x = 0. However, only one of these solutions is actually valid. To find out which one, check them both.

  • Solve:  sqrt(x + 4) + sqrt(2x – 1) = 3sqrt(x – 1)
  • This equation will also have to be squared twice. Don't forget to square that 3 in front of the square root on the right-hand side! Copyright © Elizabeth Stapel 2002-2011 All Rights Reserved

      0 = 7x^2 – 43x + 40

To solve this, use the Quadratic Formula. Then check your answers, because only one is actually valid.

  • Solve:  sqrt(x) × sqrt(x – 7) = 12
  •  

    ADVERTISEMENT

     

    This one is actually simpler than the two previous examples, because the two square roots are multiplied together, rather than added or subtracted. So this equation will need to be squared only once:

      (x + 9)(x – 16) = 0

Then the solutions are x = –9 and x = 16. But x cannot equal –9, because this would put negatives inside both radicals in the original equation. Now you check the other solution, to see if it might work.

  • Solve:  sqrt(17x – sqrt(x^2 – 5)) = 7
  • Since there is a square root inside a square root, I'll have to square twice:

      144x^2 – 833x + 1203 = 0

Using the Quadratic Formula, I get solutions of x = 401/144 and x = 3. Check these, as only one is a valid solution.

<< Previous  Top  |  1 | 2 | 3 | 4 | 5 | 6  |  Return to Index  Next >>

Cite this article as:

Stapel, Elizabeth. "Solving Radical Equations: Examples." Purplemath. Available from
    http://www.purplemath.com/modules/solverad5.htm. Accessed
 

 

 

FIND YOUR LESSON
This lesson may be printed out for your personal use.

Content copyright protected by Copyscape website plagiarism search

  Copyright 2002-2014  Elizabeth Stapel   |   About   |   Terms of Use   |   Linking   |   Site Licensing

 

 Feedback   |   Error?