# Math Help - Pulling out X to simplify, is this a correct method?

1. ## Pulling out X to simplify, is this a correct method?

Hey guys, I'm sorry for the extremely simple question, but I'm doing math by Distance Learning, and my tutor seems to be MIA. =(

So I've got the equation $f(x)=x^2-4\sqrt{x}$ and I need to find my x intercept, which i'm doing by setting $f(x)=0$.

This gives me

$0=x^2-4\sqrt{x}$

Now I'm trying to simplify it to isolate x by pulling out an x, but I'm not sure if this is a valid rule or not.

$0=x^2-4\sqrt{x}$
$0=x^2-4x^{1/2}$
$0=x(x-4(1)^{-1/2})$

If I can do this, I think that next I would have $0=x(x-4)$ which would mean that x cannot be equal to 0 or 4, and I think that these would be my x-intercept(s)?

Thoughts? I don't know why this is spiraling my brain out of control.

Also, in the original function, to find the y intercept, i plug in 0 for x right? which would give me 0, meaning that the y-intercept is through the origin? Can I use this to decide that the x-intercept is also 0?

Thanks in advance for ANY help!

2. Originally Posted by Kasper
Hey guys, I'm sorry for the extremely simple question, but I'm doing math by Distance Learning, and my tutor seems to be MIA. =(

So I've got the equation $f(x)=x^2-4\sqrt{x}$ and I need to find my x intercept, which i'm doing by setting $f(x)=0$.

This gives me

$0=x^2-4\sqrt{x}$

Now I'm trying to simplify it to isolate x by pulling out an x, but I'm not sure if this is a valid rule or not.

$0=x^2-4\sqrt{x}$
$0=x^2-4x^{1/2}$
$0=x(x-4(1)^{-1/2})$

If I can do this, I think that next I would have $0=x(x-4)$ which would mean that x cannot be equal to 0 or 4, and I think that these would be my x-intercept(s)?

Thoughts? I don't know why this is spiraling my brain out of control.

Also, in the original function, to find the y intercept, i plug in 0 for x right? which would give me 0, meaning that the y-intercept is through the origin? Can I use this to decide that the x-intercept is also 0?

Thanks in advance for ANY help!
That's the first time I've seen that tried.
And this comment does not help.

The only reason I commented -- it is the easiest way to get back to this question.

3. Originally Posted by Kasper
Hey guys, I'm sorry for the extremely simple question, but I'm doing math by Distance Learning, and my tutor seems to be MIA. =(

So I've got the equation $f(x)=x^2-4\sqrt{x}$ and I need to find my x intercept, which i'm doing by setting $f(x)=0$.

This gives me

$0=x^2-4\sqrt{x}$

Now I'm trying to simplify it to isolate x by pulling out an x, but I'm not sure if this is a valid rule or not.

$0=x^2-4\sqrt{x}$
$0=x^2-4x^{1/2}$
$0=x(x-4(1)^{-1/2})$

If I can do this, I think that next I would have $0=x(x-4)$ which would mean that x cannot be equal to 0 or 4, and I think that these would be my x-intercept(s)?

Thoughts? I don't know why this is spiraling my brain out of control.

Also, in the original function, to find the y intercept, i plug in 0 for x right? which would give me 0, meaning that the y-intercept is through the origin? Can I use this to decide that the x-intercept is also 0?

Thanks in advance for ANY help!
Hi Kasper,

$0=x^2-4\sqrt{x}$
$0=x^2-4x^{1/2}$
$0=x(x-4(1)^{-1/2})$
$x^{1/2}$ is the common factor, not $x$.

$x^{1/2}(x^{3/2}-4)=0$

$\sqrt{x}=0 \ \ or \ \ x^{3/2}-4=0$

${\color{red}x=0} \ \ or \ \ \sqrt{x^3}-4=0$

$\sqrt{x^3}=4$

$x^3=16$

${\color{red}x=\sqrt[3]{16} \ \ or \ \ x=2\sqrt[3]{2}}$

But, I believe I'd do it another way.

$x^2-4\sqrt{x}=0$

$x^2=4\sqrt{x}$

$x^4=16x$

$x^4-16x=0$

$x(x^3-16)=0$

${\color{red}x=0} \ \ or \ \ x^3=16$

${\color{red}x=\sqrt[3]{16}\ \ or \ \ x=2\sqrt[3]{2}}$

In either case, your x-intercepts are ${\color{red}0} \ \ and \ \ {\color{red}2\sqrt[3]{2}}$