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\((\sqrt{5+\sqrt5}-\sqrt{5-\sqrt5})^2=\)

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by Vincen » Mon Oct 19, 2020 9:19 am
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\((\sqrt{5+\sqrt5}-\sqrt{5-\sqrt5})^2=\)

A. \(10-4\sqrt5\)

B. \(10-2\sqrt5\)

C. \(20-8\sqrt5\)

D. \(20-4\sqrt5\)

E. \(20-2\sqrt5\)

Answer: A

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Re: \((\sqrt{5+\sqrt5}-\sqrt{5-\sqrt5})^2=\)

by Scott@TargetTestPrep » Tue Oct 27, 2020 10:06 am
Vincen wrote: ↑
Mon Oct 19, 2020 9:19 am
 \((\sqrt{5+\sqrt5}-\sqrt{5-\sqrt5})^2=\)

A. \(10-4\sqrt5\)

B. \(10-2\sqrt5\)

C. \(20-8\sqrt5\)

D. \(20-4\sqrt5\)

E. \(20-2\sqrt5\)

Answer: A

Solution:

Let A = √(5 + √5) and B = √(5 - √5) and recall that (A - B)^2 = A^2 - 2AB + B^2. So here, we have:

A^2 = 5 + √5

B^2 = 5 - √5

and

AB = √(5 + √5) * √(5 - √5) = √(25 - 5) = √20 = 2√5

Therefore, we have:

(A - B)^2 = A^2 - 2AB + B^2 = 5 + √5 - 2(2√5) + 5 - √5 = 10 - 4√5

Answer: A

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