# Thread: Should be a simple solve for X

1. ## Should be a simple solve for X

Hi, I've run into another problem on my algebra refresher. I must be doing something wrong here but I just can't see it.

$\displaystyle \sqrt{10-x}+x = 4$

Need to solve for X, of course I know the answer should be 1 but I can't seem to arrive at it.

My steps:
1: Put the lone x onto the right side.

$\displaystyle \sqrt{10-x} = 4-x$

2:Square both sides (to get rid of the square root on left side).

$\displaystyle 10-x = 16+x^2$

3: Rearrange.
$\displaystyle 0 = x^2 + x + 6$

4: At this point I could try putting this into the quadratic formula but it has a negative square root.

A mistake I may have made is in step 2, was I wrong to make the $\displaystyle x^2$ on the right hand side a positive? But even if it is left as a negative my answer comes out as...

$\displaystyle x^2 - x - 6$

Which I can make into (x-3)(x+2) I think, which would make x = 3, -2 but neither of those are the right answer of 1.

Am I doing this problem very wrong? Thanks in advance.

2. $\displaystyle (4 - x)^2 \neq 16 + x^2$.

Remember that $\displaystyle (a + b)^2 = (a + b)(a + b) = a^2 + 2ab + b^2$.

So that means $\displaystyle (4 - x)^2 = 16 - 8x + x^2$.

3. OH! I always made that same mistake when I was in high school too.

Thanks for your help, I've been staring at this for a while now.