# Math Help - Solve sqrt{x - 5} - sqrt{x - 8) = 3

1. ## Solve sqrt{x - 5} - sqrt{x - 8) = 3

$\sqrt{x-5} -\sqrt{x-8} =3$

This eventually works out to an $(A+B)^2$ = $A^2 + 2AB +B^2$ thing on the right side so that-

$x-8 + 2(\sqrt{x-8}+3) +3^2$

$\sqrt{x-5}$ = x-5 so that

x-5= x-8 + $2(\sqrt{x-8}+3) +9$

Combining like terms: -8+9 = 1; and then I subtract 1 and x from both sides to get just a -6 on the left side.

After calculating all this out I came up with 9, but 9 doesn't work. The book says that the answer is 0 with a slash through it. Does that mean no solution because 9 is the only answer and doesn't check? I also checked it with zero and came up with

$-5 +2i\sqrt{2}=3$ That can't be right... right?

I think this may be the problem: I'm thinking that my error may be in $2(\sqrt{x-8}+3)$ Does that come out to

$6\sqrt{x-8}$ which when squared =

$36(x-8)$ which = $36x-288$

Meanwhile on the left side I have 36, derived from $-6^2$ so that-

$36=-288 +36x$: Add 288 to both sides to get $324=36x$

324 divided by 36x= $9$

I realized I've skipped over some steps for the sake of brevity, so if some clarity is needed I'm happy to provide. Any and all help is greatly appreciated. Thanks.

2. ## Re: Okay, a quick question...maybe not such a quick question

$\sqrt{x-5}=3+\sqrt{x-8}$ so that $x-5=9+6\sqrt{x-8}+(x-8)$.

Do it a second time.

3. ## Re: Okay, a quick question...maybe not such a quick question

In order to isolate the radical I have to combine like terms and then move them to the other side. Right? Then, in order to elminate the radical from the left side I have to square it, no? At the same time I have to square the left side too, is that correct? And squaring $6\sqrt{x-8}$ in order to boot out the radical would seem to get me

$36(x-8)$ which will be

$36x -288$

Am I spazzing out when subtracting x from both sides?

The example in the book after getting to the point we're talking about has:

$3x+1=x+4+2\sqrt{x+4}+1$

Then it says to combine like terms so that-

$3x+1=x+5+2\sqrt{x+4}$ which gives

$2x-4=2\sqrt{x+4}$

Then it says to square both sides

$(2x-4)^2=(2\sqrt{x+4})$ which gives

$4x^2 -16x +16=4(x+4)$ so that

$4x^2 -16x +16=4x+16$ which gives

$4x^2 -20x=0$

From there you get x=0 and x=5 which I understand well enough. But where exactly am I going wrong in my OP?

$x-x=0$ so that would only leave me with a $-6$ to work with, no?

I know it's some small detail I'm missing but... I think I might have just realized it... lemme see.

No, I didn't realize it. I came up with 36(-x +9)=0. That can't be right. Moving 36 from left to right yields me the same thing but in negative. Both are false solutions but maybe they're the right false solutions?

4. ## Re: Okay, a quick question...maybe not such a quick question

You are not missing anything.
This equation has no solution.

5. ## Re: Solve sqrt{x - 5} - sqrt{x - 8) = 3

Looking at Plato's post (#2), there's no solution by not making any calculation, end of the story.