SOLUTION: How to find {{{B^2}}} in {{{1^2 + B^2}}} = {{{(sqrt(2))^2}}} ?

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Question 1025416: How to find B%5E2 in 1%5E2+%2B+B%5E2 = %28sqrt%282%29%29%5E2 ?
Answer by ikleyn(52781) About Me  (Show Source):
You can put this solution on YOUR website!
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How to find B%5E2 in 1%5E2+%2B+B%5E2 = %28sqrt%282%29%29%5E2 ?
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1%5E2+%2B+B%5E2 = %28sqrt%282%29%29%5E2   --->

1%5E2+%2B+B%5E2 = 2,

B%5E2 = 2-1,

B%5E2 = 1.

One more step, and

B = +/-1.

Notice that I edited your post.