Question

SAT Math Question: If a ^2+b^2=258, what is the value of (a+b)^2+(a-b)^2??
Please explain how you arrived at your answer. Thank you!

Answer 1

Answer:

516

Step-by-step explanation:

Let's start with the value first to see if we can use $a^2+b^2=258$ to help find it's value:

$(a+b)^2+(a-b)^2$

I'm going to use the formula $(u+v)^2=u^2+2uv+v^2$ to expand both:

$a^2+2ab+b^2+a^2-2ab+b^2$

Combining like terms:

$2a^2+2b^2$

Factoring the 2 out:

$2(a^2+b^2)$

Plug in 258 for the $a^2+b^2$:

$2(258)$

Perform the multiplication:

$516$

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Another way:

Find values for $a$ and $b$ that satisfy:

$a^2+b^2=258$

The easiest solution you might see is $a=\sqrt{258} \text{ while }b=0$.  This works because the square of $\sqrt{258}$ is 258.

So now you just plug:

$(a+b)^2+(a-b)^2$ with $a$ being $\sqrt{258}$ and $b$ being 0 into your calculator or if you are good at simplifying things without you can do that with this problem:

$(\sqrt{258}+0)^2+(\sqrt{258}-0)^2$

$(\sqrt{258})^2+\sqrt{258})^2$

$258+258$

$2(258)$

$516$

This would have worked for any pair $(a,b)$ satisfying $a^2+b^2=258$.

I wanted to show this last strategy just in case you haven't been exposed to expanding squared binomials with foil or the formula I mentioned earlier.