1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Solving an equation, I have an incorrect solution

  1. Feb 18, 2015 #1
    I have been attempting to solve the equation, I have arrived at two roots however only one of them seems to be correct. If possible could you spot a possible mistake I have made.

    ## \sqrt{3x + 10} =2 + \sqrt{x+4}##

    ##3x + 10=4+4\sqrt{x+4}+x+4##

    ##0.5x+0.5=\sqrt{x+4}##

    ##0.25x^2+0.5x+0.25=x+4##

    ##x^2+2x+1=4x+16##

    ##x^2-2x-15=0##

    ##(x-5)(x+3)=0##

    ##x=5## and ##x=-3##

    However when you put the results back into the equation only x=5 is true
    Thanks for any help
     
  2. jcsd
  3. Feb 18, 2015 #2

    Stephen Tashi

    User Avatar
    Science Advisor

    In your first step, you squared both sides of an equation. When this is done, you may create an equation that has more solutions than the original equation. ( For example, consider the equation y = 5. If we square both sides to obtain y^2 = 25 the second equation has solutions y = 5 and y = -5.)

    If you have a very sophisticated mathematical mind and some verbal skills, you can write notes in your work indicating the possible places where extra solutions may be introduced and how to intend to eliminate them. Most people don't use that approach. They simply crank out the problem and check whether the final solutions that are produced actually work in the original equation. That is what you did. The fact that not all solutions work doesn't mean you made any error in the algebraic manipulations.
     
  4. Feb 18, 2015 #3
    There are sometimes roots which do not satisfy the original equation these are called 'extraneous solutions'. Khan academy has videos explaining this far better than I ever could.

    Edit: I read equation wrong :(
     
    Last edited: Feb 18, 2015
  5. Feb 18, 2015 #4

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

     
  6. Feb 18, 2015 #5
    Yeah I realized after I posted, I edited my post so I did not to confuse anybody, I still have lots of mathematics to learn myself but I thought I could help this guy but made a mistake myself o:)
     
  7. Feb 19, 2015 #6

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    As Marcus points out, x = -3 is an "extraneous solution".

    It is a solution to the equation $$ \sqrt{3x + 10} =2 - \sqrt{x+4}\ \ .$$

    If you follow the same set of steps with this equation that you did with yours, you will eventually arrive the equation: $$0.25x^2+0.5x+0.25=x+4\ \ $$
    just as you did with your equation.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted