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. :)
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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