solve an algebraic equation involving square roots

[jusset]

I need help with the following problem:

8+ the square root of (20-x) equals 11 + the square root of (9-x)
Thanks for your help. [/list]

 

[pka]

\(\displaystyle \begin{array}{l}
8 + \sqrt {20 - x} = 11 + \sqrt {9 - x} \\
\sqrt {20 - x} - \sqrt {9 - x} = 3 \\
\left( {20 - x} \right) - 2\left( {\sqrt {20 - x} } \right)\left( {\sqrt {9 - x} } \right) + \left( {9 - x} \right) = 9 \\
20 - 2x = 2\left( {\sqrt {20 - x} } \right)\left( {\sqrt {9 - x} } \right) \\
\end{array}\).

Can you finish?
Be careful of domains.

 

[jusset]

Next I would square both sides.

I would then have 400-80x+4x^2=720-116x+4x^2
Next would be 36x =320
x= 8 2/9

Is this correct?

 

[pka]

Almost: \(\displaystyle \L \frac{80}{9}\)