Alhamdan1995 wrote:
\((\sqrt{8}+\sqrt{2})^2\) and \((\sqrt{8}-\sqrt{2})^2\)
A. 4
B. 5
C. 8
D. 10
E. 12
That's how i solved it:
1. Distributed the square: \(((\sqrt{8})^2+(2*\sqrt{8}*\sqrt{2})+(\sqrt{}2)^2+(\sqrt{8})^2-(2*\sqrt{8}*\sqrt{2})-(\sqrt{}2)^2)\)
2. Simplified, the second and third term cancel out: \(((\sqrt{8})^2+(\sqrt{8})^2) = 16\)
3. 16/2 = 8. However the correct answer is 10. Not sure what I missed.
Average of \((\sqrt{8}+\sqrt{2})^2\) and \((\sqrt{8}-\sqrt{2})^2\) =
√8+√2 = 2√2+√2 = 3√2
i.e.\( (√8+√2)^2 = (3√2)^2 = 18\)
√8-√2 = 2√2-√2 = √2
i.e. \((√8-√2)^2 = (√2)^2 = 2\)
AVerage of 18 and 2 \(= \frac{18+2}{2} = 10\)
ANswer: Option D
Alhamdan1995 Your mistake is in sign of last (√2)^2 which should be positive instead of negative
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