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If a+b=−8, and a=20/b, what is the value of a^2+b^2?

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by Brent@GMATPrepNow » Mon Nov 20, 2017 9:16 am
AAPL wrote:If $$a+b=−8$$ and $$a=20/b$$ what is the value of $$a^2+b^2$$?

A. 24
B. 32
C. 34
D. 40
E. 104

The OA is A.

Please, can any expert assist me with this PS question? I don't have it clear and I appreciate if any explain it for me. Thanks.
Given: a+b=−8 and a=20/b

Take a=20/b and multiply both sides by b to get: ab = 20

Take a + b = −8 and SQUARE both sides to get (a+b)² = (−8)²
Expand and simplify left side to get: a² + 2ab + b² = 64
Now replace ab with 20 to get: a² + 2(20)+ b² = 64
Simplify: a² + 40 + b² = 64
Subtract 40 from both sides to get: a² + b² = 24

Answer: A

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by EconomistGMATTutor » Mon Nov 20, 2017 11:05 am
If $$a+b=-8$$ and $$a=20/b$$ what is the value of $$a^2+b^2$$?

A. 24
B. 32
C. 34
D. 40
E. 104

The OA is A.

Please, can any expert assist me with this PS question? I don't have it clear and I appreciate if any explain it for me. Thanks.
Hi AAPL,
Let's take a look at your question.

$$a+b=-8$$
$$a=\frac{20}{b}$$
$$a^2+b^2=?$$

Starting from,
$$a+b=-8$$
Squaring both sides,
$$\left(a+b\right)^2=\left(-8\right)^2$$
Evaluating,
$$a^2+b^2+2ab=64...(i)$$

We know that,
$$a=\frac{20}{b}$$
Plugging value of a in (i),
$$a^2+b^2+2\left(\frac{20}{b}\right)b=64$$
$$a^2+b^2+2\left(20\right)=64$$
$$a^2+b^2+40=64$$
$$a^2+b^2=64-40$$
$$a^2+b^2=24$$

Therefore, Option A is correct.

Hope it helps.
I am available if you'd like any follow up.
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by Jeff@TargetTestPrep » Wed Nov 22, 2017 12:18 pm
AAPL wrote:If $$a+b=−8$$ and $$a=20/b$$ what is the value of $$a^2+b^2$$?

A. 24
B. 32
C. 34
D. 40
E. 104
We can square the first equation and we have:

a^2 + b^2 + 2ab = 64

Simplifying the second equation, we have:

ab = 20

Thus:

a^2 + b^2 + 40 = 64

a^2 + b^2 = 24

Answer: A

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Head of GMAT Instruction
[email protected]

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