Return to the Purplemath home page

 

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

 

Index of lessons | Purplemath's lessons in offline form |
Forums | Print this page (print-friendly version) | Find local tutors

 

Solving Radical Equations: Examples (page 3 of 6)


  • Solve the equation: sqrt(x) - 2 = 5
  • I could square both sides now, but look what I would get:

      x - 4×sqrt(x) = 21

    So, while squaring both sides at this point would not be "wrong", it would not be the most useful first step. Instead of squaring right away, I will first move the 2 over to the right-hand side, so the radical will be by itself on the left:

      sqrt(x) = 7

    Now squaring both sides will work better:

      x = 49

    Checking, I get:

      5 = 5

    So the solution is x = 49.

  • Find the solution:
    • sqrt(x - 3) - sqrt(x) = 3

    This problem is a bit more messy than the previous two. I cannot get the radical by itself on one side, because there are two radicals. So how can I solve this algebraically? By squaring both sides twice. Here's what it looks like:

      4 = x

    Warning: Do NOT try to do these steps in your head. Take the time to write things out completely, so you won't make mistakes!

     

    ADVERTISEMENT

     

    Checking my solution, I get:

      -1 does not equal 3

    Hmm... According to this, there is no solution. I'll check the graph of the two lines:

      y = sqrt(x - 3) - sqrt(x),  y = 3

    ...to see if it looks like there ought to be a solution:

      graph of system

    No; according to the graph, it does not appear that these lines intersect (and calculus techniques can prove this). Why did it appear that there was a solution? Look at the graphs from the second squaring: Copyright © Elizabeth Stapel 2002-2011 All Rights Reserved

      y = x2 – 12x + 36
      y = x2 – 3x

      graph of squared equations

    So I came up with an algebraic solution because I had accidentally created one by my repeated squaring. But that "solution" didn't check out in the original equation, so the actual answer is that there is no solution.

On the other hand, look at the following...

<< 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/solverad3.htm. Accessed
 

 

 

MATHHELP LESSONS
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?