1. ## Square root equation

Not sure where to even star.

2. ## Re: Square root equation

Maybe square both sides X3

3. ## Re: Square root equation

I would try squaring both sides first...

4. ## Re: Square root equation

$1+\sqrt{x+\sqrt{2x+1}}=5+\sqrt{x}$

$(\sqrt{x+\sqrt{2x+1}})^2=(4+\sqrt{x})^2$

$x+\sqrt{2x+1}=16+8\sqrt{x}+x$

$(\sqrt{2x+1})^2=(16+8\sqrt{x})^2$

$2x+1=256+128\sqrt(x)+64x$

$128\sqrt{x}+62x+255=0$

Then I'm stuck.

5. ## Re: Square root equation

Originally Posted by Remriel
$128\sqrt{x}+62x+255=0$

Then I'm stuck.

The "128" coefficient should be "256." Check your expansion.

Let $y = \sqrt{x}$ (we'll assume $x \ge 0$...if $x < 0$ then...).

6. ## Re: Square root equation

$256\sqrt{x}+62x+255=0$

$62y^2+256y+255=0\\\\ y=\frac{-256\pm\sqrt{2296}}{124}$

$y=-\frac{128\pm \sqrt{574}}{62}$

7. ## Re: Square root equation

Looks like no real solutions

8. ## Re: Square root equation

Originally Posted by pickslides
Looks like no real solutions
That is correct. See this.

9. ## Re: Square root equation

Figures. Thanks.