Equation with square root

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  • Thread starter magda21
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  • #1
magda21
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Please Help me solve it \[ \sqrt{x}+\sqrt{x+8}=8 \] thanks
 

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  • #2
MarkFL
Gold Member
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11
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?
 
  • #3
magda21
3
0
x+8=64-x
2x=56
x=28
what I'm doing wrong?
 
  • #4
MarkFL
Gold Member
MHB
13,302
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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?
 
  • #5
magda21
3
0
\[ \sqrt{x}=\frac{7}{2} \]
x=\frac{49}{4}
I see, thank you so much
 
  • #6
MarkFL
Gold Member
MHB
13,302
11
Yes, and once we verifiy is works in the original equation, which it does, then we're done. :)
 

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