- #1

magda21

- 3

- 0

Please Help me solve it \[ \sqrt{x}+\sqrt{x+8}=8 \] thanks

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- Thread starter magda21
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- #1

magda21

- 3

- 0

Please Help me solve it \[ \sqrt{x}+\sqrt{x+8}=8 \] thanks

- #2

MarkFL

Gold Member

MHB

- 13,302

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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?

2x=56

x=28

what I'm doing wrong?

- #4

MarkFL

Gold Member

MHB

- 13,302

- 11

\(\displaystyle \sqrt{x+8}=8-\sqrt{x}\)

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

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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