Answer: 200
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Work Shown:
One approach is to simplify the stuff under the second power exponent. After that simplification, you would then square the result.
![x = √(2)+√(8)+√(18)+√(32)\\\\x = √(2)+√(4*2)+√(9*2)+√(16*2)\\\\x = √(2)+√(4)*√(2)+√(9)*√(2)+√(16)*√(2)\\\\x = 1√(2)+2√(2)+3√(2)+4√(2)\\\\x = (1+2+3+4)√(2)\\\\x = 10√(2)\\\\x^2 = \left(10√(2)\right)^2\\\\x^2 = \left(10√(2)\right)*\left(10√(2)\right)\\\\x^2 = 10*10√(2)*√(2)\\\\x^2 = 100√(2*2)\\\\x^2 = 100√(4)\\\\x^2 = 100*2\\\\x^2 = 200\\\\](https://img.qammunity.org/2021/formulas/mathematics/high-school/4k5yybvd0j8huyfl5qbutttw1ofk4c70zu.png)
So,
![\left(√(2)+√(8)+√(18)+√(32)\right)^2 = 200](https://img.qammunity.org/2021/formulas/mathematics/high-school/skhwfc8dctyv2qg7rwc5x0d59c6mev9sqv.png)
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Checking the answer:
Using a calculator,
sqrt(2)+sqrt(8)+sqrt(18)+sqrt(32) = 14.142135623731
Then you square that to get (14.142135623731)^2 = 200.000000000001
Your calculator may not have that 1 at the end. It shouldn't be there and it's a result of rounding error. But it's close enough to 200.