Square roots

a.nourrice · **Joined:** Tue, 8 May 2012 19:30:50 UTC **Posts:** 4

Hi, I am working on a problem set and I have reached at the following:

((√c-8)/√c)((4√c)/(√c-24))= -1/4

4((√c-8)/(√c-24))= -1/4

How do i solve for c?

Denis · **Posted:** Tue, 8 May 2012 20:47:51 UTC

a.nourrice wrote:

(√c-8)

Unclear: is that SQRT(c) - 8 or SQRT(c-8)?

a.nourrice · **Joined:** Tue, 8 May 2012 19:30:50 UTC **Posts:** 4

its SQRT(c) - 8

Justin · **Joined:** Wed, 21 May 2003 04:27:18 UTC **Posts:** 990

a.nourrice wrote:

Hi, I am working on a problem set and I have reached at the following:

((√c-8)/√c)((4√c)/(√c-24))= -1/4

4((√c-8)/(√c-24))= -1/4

How do i solve for c?

Another question: Is the second line your attempt to simplify the first one?

a.nourrice · **Joined:** Tue, 8 May 2012 19:30:50 UTC **Posts:** 4

yes, the second equation is a simplification of the first one which I have done... I am not sure if it is the right step to solve for c, I am confused because I am led to believe i need to apply rationalization here

Justin · **Joined:** Wed, 21 May 2003 04:27:18 UTC **Posts:** 990

a.nourrice wrote:

yes, the second equation is a simplification of the first one which I have done... I am not sure if it is the right step to solve for c, I am confused because I am led to believe i need to apply rationalization here

Assuming you mean $$4\frac{\sqrt{c} - 8}{\sqrt{c} - 24} = -\frac{1}{4}$ :

Forget rationalizing, just cross multiply:

$16\sqrt{c} - 128 = 24 - \sqrt{c} \implies 17\sqrt{c} = 152$

Can you take it from here?

a.nourrice · **Joined:** Tue, 8 May 2012 19:30:50 UTC **Posts:** 4

Yes! I can take over from where you left. Thank you for your help

Justin · **Joined:** Wed, 21 May 2003 04:27:18 UTC **Posts:** 990

You are welcome!

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