MHB How to Solve an Equation with Square Roots?

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To solve the equation \(\sqrt{x}+\sqrt{x+8}=8\), first rearrange it to \(\sqrt{x+8}=8-\sqrt{x}\). Squaring both sides gives \(x+8=64-16\sqrt{x}+x\), leading to the equation \(16\sqrt{x}=56\). Dividing by 8 simplifies this to \(2\sqrt{x}=7\), resulting in \(\sqrt{x}=\frac{7}{2}\) and \(x=\frac{49}{4}\). Verification shows that this solution satisfies the original equation.
magda21
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Please Help me solve it \[ \sqrt{x}+\sqrt{x+8}=8 \] thanks
 
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Hello, and welcome to MHB! :)

I would begin by arranging as follows:

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

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?
 
x+8=64-x
2x=56
x=28
what I'm doing wrong?
 
Ah, you are not squaring the RHS correctly. Let's go back to:

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

Now, recall that:

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

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

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

Collecting like terms, we can arrange this as:

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

Divide through by 8:

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

Can you proceed?
 
\[ \sqrt{x}=\frac{7}{2} \]
x=\frac{49}{4}
I see, thank you so much
 
Yes, and once we verifiy is works in the original equation, which it does, then we're done. :)
 
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