If $a^2 + b^2 = 4$ and $(a-b)^2 =2$ , what is the value of $ab$ ?

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Answer 1 · 2018-03-30T01:07:24

The value of $ab$ is $1$ .

We can expand $(a- b)^2$ to see that $(a- b)^2 = a^2 - 2ab + b^2$ , thus $a^2 + b^2 - 2ab = 2$ .

It follows that $a^2 + b^2 = 2 + 2ab$ , and since we're given that $a^2+ b^2 = 4$ , we can say $4 = 2 + 2ab$ . Doing a little algebra we get

$4 = 2(1 + ab)$

$2 = 1 + ab$

$ab = 1$

Hopefully this helps!

If a^2 + b^2 = 4a^2 + b^2 = 4 and (a-b)^2 =2(a-b)^2 =2, what is the value of abab?