# Equation with square root

• MHB
magda21
Please Help me solve it $\sqrt{x}+\sqrt{x+8}=8$ thanks

Gold Member
MHB
Hello, and welcome to MHB! :)

I would begin by arranging as follows:

$$\displaystyle \sqrt{x+8}=8-\sqrt{x}$$

I really just want to arrange it so that there isn't two radicals on the same side. Now, what do you get when you square both sides?

magda21
x+8=64-x
2x=56
x=28
what I'm doing wrong?

Gold Member
MHB
Ah, you are not squaring the RHS correctly. Let's go back to:

$$\displaystyle \sqrt{x+8}=8-\sqrt{x}$$

Now, recall that:

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

And so, when we square both sides of our equation (bearing in mind that we must check for extraneous solutions), we get:

$$\displaystyle x+8=64-16\sqrt{x}+x$$

Collecting like terms, we can arrange this as:

$$\displaystyle 16\sqrt{x}=56$$

Divide through by 8:

$$\displaystyle 2\sqrt{x}=7$$

Can you proceed?

magda21
$\sqrt{x}=\frac{7}{2}$
x=\frac{49}{4}
I see, thank you so much

Gold Member
MHB
Yes, and once we verifiy is works in the original equation, which it does, then we're done. :)