# Math Help - Irrational equations

1. ## Irrational equations

I'm almost in tears, I've been trying to solve this equation yesterday and today with no success. Please help!

My attempt at solving (I'm going to put numbers by every line so you can help me find my mistakes easier):
1)
2)
3)
4)
5)
6)
7)
8)
9)

2. You didn't expand $\left(8+2\sqrt{8}\sqrt{x-8}+x-8\right)^2$ correctly.

3. But how do I do that? Is there a specific formula?

4. ... and you can make the calculation a lot easier by subtracting x + 8 from both sides of line (3) before squaring it.

5. Originally Posted by Evaldas
But how do I do that? Is there a specific formula?
Well, first leave the 8's behind. No reason to carry them forward.
So you have to expand $\left(2\sqrt{8}\sqrt{x-8}+x\right)^2$.
For this, you sure know that $(a+b)^2 = a^2+2ab+b^2$, right?

On the right hand side of the equation you have a positive 8, and x, and a negative 8 that are all not under square roots. Move them to the left side of the equation. You will be left with 8 on the left side of the equation. Now square both sides and the square roots will cancel out on the right side. Solve for x.

7. Originally Posted by Evaldas
$\sqrt{x+8} - \sqrt{x-8} = \sqrt{8}$

square both sides ...

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

combine like terms and radicals on the left ...

$2x - 2\sqrt{x^2-64} = 8$

divide all terms by 2 ...

$x - \sqrt{x^2-64} = 4$

rearrange equation to move radical to the right side ...

$x - 4 = \sqrt{x^2 - 64}$

square both sides ...

$x^2 - 8x + 16 = x^2 - 64$

combine like terms ...

$-8x = -80$

$x = 10$

8. Thank you all!