1. ## Help solving for x please... 2(x^2 -1)^(1/5)

Flabbergasted as to how to go about doing this... Would really appreciate some help. I've only ever been asked to do this with quadratic and cubic polynomials.

$\displaystyle f(x) = 2(x^2 -1)^\frac{1}{5}$

Edit: Oops, yeah, I want to solve for f(x) = 0. I'm trying to find the x-intercepts.

2. Originally Posted by liquidwater
Flabbergasted as to how to go about doing this... Would really appreciate some help. I've only ever been asked to do this with quadratic and cubic polynomials.

$\displaystyle f(x) = 2(x^2 -1)^\frac{1}{5}$
Dear liquidwater,

Do you mean you want to subject x in this equation???

3. Originally Posted by liquidwater
Flabbergasted as to how to go about doing this... Would really appreciate some help. I've only ever been asked to do this with quadratic and cubic polynomials.

$\displaystyle f(x) = 2(x^2 -1)^\frac{1}{5}$
You need an equation in order to solve for $\displaystyle x$. Is this equal to $\displaystyle 0$?

Anyway, if you want to factorise it...

$\displaystyle f(x) = 2(x^2 - 1)^{\frac{1}{5}}$

$\displaystyle = 2[(x - 1)(x + 1)]^{\frac{1}{5}}$

$\displaystyle = 2(x - 1)^{\frac{1}{5}}(x + 1)^{\frac{1}{5}}$.

4. Originally Posted by liquidwater
Flabbergasted as to how to go about doing this... Would really appreciate some help. I've only ever been asked to do this with quadratic and cubic polynomials.

$\displaystyle f(x) = 2(x^2 -1)^\frac{1}{5}$
Dear liquidwater,

Do you mean you want to subject x in this equation???

5. Originally Posted by Sudharaka
Dear liquidwater,

Do you mean you want to subject x in this equation???
Thanks, I updated my original post.

Prove It, thanks... I still can't figure how to find f(x) = 0 with that.

6. $\displaystyle 2(x^2 - 1)^{\frac{1}{5}} = 0$

$\displaystyle 2[(x - 1)(x + 1)]^{\frac{1}{5}} = 0$

$\displaystyle 2(x - 1)^{\frac{1}{5}}(x + 1)^{\frac{1}{5}} = 0$.

So $\displaystyle (x - 1)^{\frac{1}{5}} = 0$ or $\displaystyle (x + 1)^{\frac{1}{5}} = 0$.

Case 1:

$\displaystyle (x - 1)^{\frac{1}{5}} = 0$

$\displaystyle x - 1 = 0$

$\displaystyle x = 1$.

Case 2:

$\displaystyle (x + 1)^{\frac{1}{5}} = 0$

$\displaystyle x + 1 = 0$

$\displaystyle x = -1$.

7. Originally Posted by Prove It
$\displaystyle 2(x^2 - 1)^{\frac{1}{5}} = 0$

$\displaystyle 2[(x - 1)(x + 1)]^{\frac{1}{5}} = 0$

$\displaystyle 2(x - 1)^{\frac{1}{5}}(x + 1)^{\frac{1}{5}} = 0$.

So $\displaystyle (x - 1)^{\frac{1}{5}} = 0$ or $\displaystyle (x + 1)^{\frac{1}{5}} = 0$.

Case 1:

$\displaystyle (x - 1)^{\frac{1}{5}} = 0$

$\displaystyle x - 1 = 0$

$\displaystyle x = 1$.

Case 2:

$\displaystyle (x + 1)^{\frac{1}{5}} = 0$

$\displaystyle x + 1 = 0$

$\displaystyle x = -1$.
Thanks a lot.

Can the $\displaystyle ^\frac{1}{5}$ be discarded because we will be making x = 0, 0 to the power of anything is just going to be 0?

And $\displaystyle 2*0$ is just going to be 0... so the 2 is discarded?

(if I'm making sense).

8. Originally Posted by liquidwater
Thanks a lot.

Can the $\displaystyle ^\frac{1}{5}$ be discarded because we will be making x = 0, 0 to the power of anything is just going to be 0?

And $\displaystyle 2*0$ is just going to be 0... so the 2 is discarded?

(if I'm making sense).
In this case you could, but if the RHS wasn't 0, you'd have to take both sides to the power of 5.

9. Originally Posted by Prove It
In this case you could, but if the RHS wasn't 0, you'd have to take both sides to the power of 5.
Thanks so much, I understand

10. You can, of course, take the fifth power here:

$\displaystyle 2(x^2- 1)^{1/5}= 0$

Divide both sides by 2: $\displaystyle (x^2- 1)^{1/5}= 0/2= 0$

Take the fifth power of both sides: $\displaystyle x^2- 1= 0$

Add 1 to both sides: $\displaystyle x^2= 1$

So $\displaystyle x= \pm 1$.

More generally, if $\displaystyle 2(x^2- 1)^{1/5}= a$,
for a any number, $\displaystyle (x^2- 1)^{1/5}= \frac{a}{2}$
$\displaystyle x^2- 1= \left(\frac{a}{2}\right)^5= \frac{a^5}{32}$
$\displaystyle x^2= 1+ \frac{a^5}{32}$

$\displaystyle x= \pm\sqrt{1+ \frac{a^5}{32}}$