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What is the average (arithmetic mean) of (\sqrt{8}+\sqrt{2})^2
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15 Jun 2026, 03:35
15 Jun 2026, 03:35
Expert Reply
00:00
Question Stats:
75% (01:02) correct
25% (01:34) wrong based on 8 sessions
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What is the average (arithmetic mean) of \((\sqrt{8}+\sqrt{2})^2\) and \((\sqrt{8}-\sqrt{2})^2\) ?
A. 4
B. 5
C. 8
D. 10
E. 12
A. 4
B. 5
C. 8
D. 10
E. 12
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What is the average (arithmetic mean) of (\sqrt{8}+\sqrt{2})^2
[#permalink]
22 Jun 2026, 07:52
22 Jun 2026, 07:52
Expert Reply
Explanation
Using \((a+b)^2=a^2+2ab+b^2:\)
\((\sqrt{8}+\sqrt{2})^2=8+2\sqrt{8}\sqrt{2}+2=10+2\sqrt{16}=10+8=18\)
\((\sqrt{8}−\sqrt{2})^2=8−2\sqrt{8}\sqrt{2}+2=10−8=2\)
Now take the average:
\(\frac{18+2}{2}\)=\(\frac{20}{2}=10.\)
Answer: D
Using \((a+b)^2=a^2+2ab+b^2:\)
\((\sqrt{8}+\sqrt{2})^2=8+2\sqrt{8}\sqrt{2}+2=10+2\sqrt{16}=10+8=18\)
\((\sqrt{8}−\sqrt{2})^2=8−2\sqrt{8}\sqrt{2}+2=10−8=2\)
Now take the average:
\(\frac{18+2}{2}\)=\(\frac{20}{2}=10.\)
Answer: D
gmatclubot
What is the average (arithmetic mean) of (\sqrt{8}+\sqrt{2})^2 [#permalink]
22 Jun 2026, 07:52
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