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The function g(x)=√(5+x)-x

  1. Oct 20, 2015 #1
    1. The problem statement, all variables and given/known data
    g(x)=√(5+x)-x

    a) Write down D(g)

    b) Solve the equation
    √(5+x)-x=3
    2. Relevant equations
    √(5+x)-x=3

    3. The attempt at a solution
    a)
    x∈ℝ: x>-5
    b)

    √(5+x)-x=3
    √(5+x)=x+3
    (5+x)=(x+3)^2
    (5+x)=(x+3)(x+3)
    (5+x)=x^2+6x+9
    x^2+5x+4=0
    (x+4)(x+1)
    x=-4 or x=-1

    √(5-4)--4=3
    √(1)+4=3 False

    √(5-1)--1=3
    2+1=3 True
    x=-1
     
    Last edited: Oct 20, 2015
  2. jcsd
  3. Oct 20, 2015 #2
    you can't take sqroot of 1 -1. sqrt(x^2)=|x|
     
  4. Oct 20, 2015 #3
    Seems right. You haven't done a) though.
     
  5. Oct 20, 2015 #4
    Oops, I've added my answer,

    Thank you,
     
  6. Oct 20, 2015 #5
    That's incorrect.(Assuming that you have the condition that the output must be real.) If x were to be less than -5, then you would have to take square roots of negative numbers, so your output would no longer be real.
     
  7. Oct 20, 2015 #6

    RUber

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    Homework Helper

    For a) you wrote x<-5, that is the opposite of the domain.
    Your solution was x = -1. That is not in x<-5.
     
  8. Oct 20, 2015 #7

    Mark44

    Staff: Mentor

    I don't know what you meant, but your first sentence above is incorrect. 1 - 1 = 0, and ##\sqrt{0} = 0##.
     
  9. Oct 20, 2015 #8
    I meant to say x>-5 so the root will not be of a negative number.
     
  10. Oct 20, 2015 #9
    Yup, that's right.
     
  11. Oct 20, 2015 #10
    Thank you,
     
  12. Oct 20, 2015 #11

    ehild

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    Homework Helper
    Gold Member


    You wrote that x should be less than -5. Can you take the square root of x-5 then? So what is the correct domain of g(x)?
    b)
    √(5+x)-x=3
    √(5+x)=x+3
    (5+x)=(x+3)^2
    (5+x)=(x+3)(x+3)
    (5+x)=x^2+6x+9
    x^2+5x+4=0
    (x+4)(x+1)
    x=-4 or x=-1[/QUOTE]
    Correct.
    Correct.
    Is x=-1 inside the domain?
     
  13. Oct 20, 2015 #12
    I did fix this it was a typo

    Yes it is
     
  14. Oct 20, 2015 #13
    I
    meant sqrt of 1 doesn't equal -1
     
  15. Oct 20, 2015 #14

    SammyS

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    Staff Emeritus
    Science Advisor
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    Gold Member

    The domain is not quite right. ##\ \sqrt{ 0 } \ ## is a real number, so -5 is in the domain.
     
  16. Oct 20, 2015 #15
    You're right. I overlooked that fact. The inequality should be weak @Jaco Viljoen
     
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