1. Limited time only! Sign up for a free 30min personal 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!

Homework Help: Factoring question

  1. Jan 31, 2010 #1
    1. The problem statement, all variables and given/known data

    I need to find the point at which these functions intersect:
    y = -sqrt(x)
    y = x - 6


    2. Relevant equations



    3. The attempt at a solution

    I set them to equate:
    -sqrt(x) = x - 6
    -sqrt(x) - x + 6 = 0

    Now how do I find the root?
     
  2. jcsd
  3. Jan 31, 2010 #2
    equate them tgt , [tex]\sqrt{x}=6-x[/tex] .
    Now square both sides , x=36+x²-12x ... continue from here.
     
  4. Jan 31, 2010 #3
    Do I move them all to one side and get x^2 -13x + 36 = 0?
     
  5. Feb 1, 2010 #4

    Mark44

    Staff: Mentor

    Yes. Now factor.

    Be sure to check each solution in the original equation, though. When you square both sides of an equation, extraneous solutions are sometimes introduced, values that are not solutions of your original equation.
     
  6. Feb 1, 2010 #5

    ideasrule

    User Avatar
    Homework Helper

    Alternatively, you can say u=sqrt(x) and rewrite the equation as -u-u^2+6=0
     
  7. Feb 1, 2010 #6
    Okay nice.
    So I can't use x = 9 because -sqrt(9) not= 9-6, correct?
     
  8. Feb 1, 2010 #7
    Ah okay, that u substitution seems easier:
    Let u = sqrt(x), then
    u^2 + u - 6 = 0
    (u+3)(u-2) = 0
    u = -3, u = 2
    sqrt(x) = -3, sqrt(x) = 2
    x = 9, x = 4
     
  9. Feb 1, 2010 #8
    Thanks guys!
     
  10. Feb 1, 2010 #9

    Mark44

    Staff: Mentor

    But x = 9 is not a solution of sqrt(x)= 6 -x, which is what you started with. x = 9 is an extraneous solution that I warned you of.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook