# Math Help - Someone please rearrange this for me

1. ## Someone please rearrange this for me

Hello there,
Usually I'm quite good at the whole rearranging process, but my answer for one question doesn't fit with what the book says it should be. What's more, I think I've found a mistake in their rearranging.
Could someone please rearrange this for a single X value, and tell me what they get?
Thank you.

2. Hi
Originally Posted by Naur
Hello there,
Usually I'm quite good at the whole rearranging process, but my answer for one question doesn't fit with what the book says it should be. What's more, I think I've found a mistake in their rearranging.
Could someone please rearrange this for a single X value, and tell me what they get?
Thank you.
$
\frac{x}{3\sqrt{4+x^2}}=\frac{1}{5}$

Square both sides and use $\sqrt{u}^2=|u|$ : $\frac{x^2}{9\sqrt{4+x^2}^2}=\frac{1}{25}\implies \frac{x^2}{9|4+x^2|}=\frac{1}{25}$

As $x^2+4 > 0, \,|x^2+4|=x^2+4$ thus the equation becomes $\frac{x^2}{9(4+x^2)}=\frac{1}{25}$

Multiply both sides by $25\times 9(4+x^2)$ : $\frac{25 x^2 \times 9(x^2+4)}{9(x^2+4)}=\frac{25\times 9(x^2+4)}{25}$

Simplify : $25 x^2 =9(x^2+4)$

and from this you should be able to find the possible values of $x$.

EDIT : You'll get two values for $x$ but there is only one which is the right one. Try to find out why.

3. Oh thank you! I was so convinced the book was lying to me. I forgot that
x/1/25 is equal to 25x. Although I used a much simpler method than you did, thank you.
I only get one x-value though. It's 1.5, which is correct, but it's odd that you say I should be getting two.

I have another maths problem, I'll post it in a new thread in a second. Thanks very much for your help.

4. Originally Posted by Naur
I only get one x-value though. It's 1.5, which is correct, but it's odd that you say I should be getting two.
The last equation I wrote gives $x=\pm\sqrt{\text{something}}$ and the right solution is the positive one since $x=\frac{3\sqrt{4+x^2}}{5}>0$.