Find the zeros of the function:

[JBauer]

f(x) = third square root of |x^2 - 9| - 3

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

Thanks,

J.B.

 

[MathStudent]

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

 

[hypermorphism]

Do you mean
[tex]\sqrt{\sqrt{\sqrt{|x^2 - 9| - 3}}}[/tex]
or
[tex](|x^2 - 9| - 3)^{\frac{1}{3}}[/tex]
?

 

[stunner5000pt]

So

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)

 

[JBauer]

Let me make sure I got the equation right
[tex] f(x) = \sqrt[3]{\mid x^2 - 9 \mid - 3} [/tex]

Phew... took me a while but I got the Latex

This equation is correct, although the - 3 is not under the radical.

Thanks for the help,

J.B.

 

[JBauer]

So

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)

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...

 

[newcool]

[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.

 

[stunner5000pt]

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...

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.

 

[poolwin2001]

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.

|x^2 - 9| is allways >0
What he may have meant was take cases x^2>=9 and x^2<9

 

[stunner5000pt]

|x^2 - 9| is allways >0
What he may have meant was take cases x^2>=9 and x^2<9

my mistake, that's what i meant the argument of the absolute value

 

[HallsofIvy]

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"?