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Homework Help: Find the zeros of the function:

  1. Jan 19, 2005 #1
    f(x) = third square root of |x^2 - 9| - 3

    I need set everything = to 0, but then what? I'm stuck.


  2. jcsd
  3. Jan 19, 2005 #2
    Let me make sure I got the equation right
    [tex] f(x) = \sqrt[3]{|x^2 - 9|} - 3 [/tex]

    Phew... took me a while but I got the Latex :yuck:
    Last edited: Jan 19, 2005
  4. Jan 19, 2005 #3
    Do you mean
    [tex]\sqrt{\sqrt{\sqrt{|x^2 - 9| - 3}}}[/tex]
    [tex](|x^2 - 9| - 3)^{\frac{1}{3}}[/tex]
  5. Jan 19, 2005 #4

    firstly rearrange it to make it look nicer(according to me at least)

    [tex] \sqrt[3]{|x^2 - 9|} = 3 [/tex]

    try and remove the cube root sign first

    then consider teh cases when the |x^2 - 9| is greater than and lesser than or equal to zero

    in the first case you simply drop teh absolute values

    inthe second case you place a negatie sign in front of the absolute value term and drop the absolute value signs

    then solve for x, you should get 4 answers, 2 real answers, and 2 complex (square root of 1, related)
  6. Jan 19, 2005 #5
    This equation is correct, although the - 3 is not under the radical.

    Thanks for the help,

  7. Jan 19, 2005 #6
    Ok, I have the cube root sign removed and my current equation is:

    |x^2 - 9| = 27

    although I do not remember how to work out the absolute values from here...
  8. Jan 19, 2005 #7
    [tex]x^2 - 9 = 27, x =+- 6[/tex]
    [tex]x^2 - 9 = -27[/tex] is false because absolute value can't equal a negative number.
  9. Jan 20, 2005 #8
    ok when you have aboslute values you have consider the arguemnt (stuff inside the absaolute value) to be greater than zero and lesser than or equal to zero.

    so first you'll have |x^2 - 9| > 0, here you should simply keep the whole expression as positive, drop the absolute values and then solve with this equated to 27.

    which will give you

    x^2 - 9 = 27, and solve.

    and secondly, you'll have |x^2-9|<=0 in this case you have to take the whole expression to be negative. That is drop the absolute values, and put a negative sign in front of the the expression

    which willl give you

    - (x^2 - 9) = 27 , and solve.
  10. Jan 20, 2005 #9
    |x^2 - 9| is allways >0 :biggrin:
    What he may have meant was take cases x^2>=9 and x^2<9
  11. Jan 20, 2005 #10
    my mistake, thats what i meant the argument of the absolute value
  12. Jan 20, 2005 #11


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    The "|x2-9|= -27" was wrong but the point that either
    x2-9= 27 or x2- 9= -27 is still true.

    Of course, x2- 9= -27 is equivalent to x2= -18 which can't be true for x any real number.

    By the way, am I the only one who is infuriated by "the third square root of"?

    My first reaction was "I thought numbers only had two square roots!

    Oh, you mean cube root!"

    Isn't "cube root" (or just "third root") easier to write than "third square root"?
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