# Thread: Solving for x

1. ## Solving for x

Something has me confused :

q^2 = u^2 + x^2 Make x the subject

I got x= squareroot a^2 - u^2 as the answer.

Having looked at the solution it's telling me the answer should be x = squareroot a^2 = u^2 ......is this their mistake or my mistake as usual?

Thanks

2. Originally Posted by komodo
Something has me confused :

q^2 = u^2 + x^2 Make x the subject

I got x= squareroot a^2 - u^2 as the answer.

Having looked at the solution it's telling me the answer should be x = squareroot a^2 = u^2 ......is this their mistake or my mistake as usual?

Thanks
The correct solution is $\displaystyle x = \pm \sqrt{a^2 - u^2}$ as you say, though note the +/-. (I'm assuming the "q" in the original equation should have been an "a.") I am guessing the given solution is supposed to be this and the second = sign is meant to be a minus.

-Dan

3. Originally Posted by komodo
Something has me confused :

q^2 = u^2 + x^2 Make x the subject

I got x= squareroot a^2 - u^2 as the answer.

Having looked at the solution it's telling me the answer should be x = squareroot a^2 = u^2 ......is this their mistake or my mistake as usual?

Thanks
$\displaystyle x=\sqrt{a^2-u^2}, x=- \sqrt{a^2-u^2}$

4. Originally Posted by topsquark
The correct solution is $\displaystyle x = \pm \sqrt{a^2 - u^2}$ as you say, though note the +/-. (I'm assuming the "q" in the original equation should have been an "a.") I am guessing the given solution is supposed to be this and the second = sign is meant to be a minus.

-Dan
Yeah the q should have been an a.....so I'm guessing the = sign is just a typo on their behalf and it should have been a minus sign.

Originally Posted by abhishekkgp
$\displaystyle x=\sqrt{a^2-u^2}, x=- \sqrt{a^2-u^2}$
Thanks