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

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

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Answer 1 · 2017-01-23T00:45:17

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

Explanation:

Given that:

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

Divide both sides by $2$ $2$ to find:

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

Hence:

$a + b = \pm \sqrt{32} = \pm \sqrt{4^{2} \cdot 2} = \pm 4 \sqrt{2}$ $a+b = +-sqrt(32) = +-sqrt(4^2*2) = +-4sqrt(2)$

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Bonus

Given:

$a - b = 10$ $a-b=10$

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

Adding these two equations, we find:

$2 a = 10 \pm 4 \sqrt{2}$ $2a = 10+-4sqrt(2)$

So $a = 5 \pm 2 \sqrt{2}$ $a = 5+-2sqrt(2)$

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

That is, the two possible solutions are:

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

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

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

1 Answer

Explanation:

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

1 Answer

Explanation:

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