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

**Kasper** · May 28th 2009, 10:14 AM

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!

**aidan** · May 28th 2009, 11:33 AM

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.

**masters** · May 28th 2009, 11:57 AM

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 ${\color{red}0} \ \ and \ \ {\color{red}2\sqrt[3]{2}}$

Your y-intercept is indeed 0.

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