If $a-b=10$ and $2*(a+b)^2 = 64$ then how do you find $a+b$ ?

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Answer 1 · 2017-01-23T00:45:17

$a+b = +-4sqrt(2)$

Given that:

$2*(a+b)^2 = 64$

Divide both sides by $2$ to find:

$(a+b)^2 = 32$

Hence:

$a+b = +-sqrt(32) = +-sqrt(4^2*2) = +-4sqrt(2)$

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Bonus

Given:

$a-b=10$

$a+b=+-4sqrt(2)$

Adding these two equations, we find:

$2a = 10+-4sqrt(2)$

So $a = 5+-2sqrt(2)$

Then $b = a-10 = -5+-2sqrt(2)$

That is, the two possible solutions are:

$(a, b) = (5+2sqrt(2), -5+2sqrt(2))$

$(a, b) = (5-2sqrt(2), -5-2sqrt(2))$

If a-b=10a-b=10 and 2*(a+b)^2 = 642*(a+b)^2 = 64 then how do you find a+ba+b ?