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

Jan 23, 2017

$a + b = \pm 4 \sqrt{2}$

Explanation:

Given that:

$2 \cdot {\left(a + b\right)}^{2} = 64$

Divide both sides by $2$ to find:

${\left(a + b\right)}^{2} = 32$

Hence:

$a + b = \pm \sqrt{32} = \pm \sqrt{{4}^{2} \cdot 2} = \pm 4 \sqrt{2}$

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Bonus

Given:

$a - b = 10$

$a + b = \pm 4 \sqrt{2}$

Adding these two equations, we find:

$2 a = 10 \pm 4 \sqrt{2}$

So $a = 5 \pm 2 \sqrt{2}$

Then $b = a - 10 = - 5 \pm 2 \sqrt{2}$

That is, the two possible solutions are:

$\left(a , b\right) = \left(5 + 2 \sqrt{2} , - 5 + 2 \sqrt{2}\right)$

$\left(a , b\right) = \left(5 - 2 \sqrt{2} , - 5 - 2 \sqrt{2}\right)$