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What is the solution of [(7 + √48)∧1/2 + (7 – √48)…

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by user123321 » Tue Oct 25, 2011 11:43 am
[email protected] wrote:What is the solution of [(7 + √48)∧1/2 + (7 – √48)∧1/2]∧2 ?
(A) 1
(B) 7 – 4√3
(C) 14 – 4√3
(D) 14
(E) 16
7+√48 = ( √4 + √3 )^2
and
7-√48 = ( √4 - √3 )^2

substituting in given question we get
(2√4)^2 = 16

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by smackmartine » Tue Oct 25, 2011 11:46 am
IMO E
[(7 + √48)∧1/2 + (7 - √48)∧1/2]∧2

=[(7 + √48)∧1/2]^2 + [(7 - √48)∧1/2]^2 + 2 [(7 + √48)*(7 - √48)]^1/2

=14+ 2[7^2 - (√48)^2]^1/2 {as (a+b)(a-b)=a^2 - b^2}
=14 + 2[49-48]^1/2
=14+2
=16
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by GMATGuruNY » Tue Oct 25, 2011 12:41 pm
[email protected] wrote:What is the solution of [√(7 + √48) + √(7 – √48)]² ?
(A) 1
(B) 7 – 4√3
(C) 14 – 4√3
(D) 14
(E) 16
We can reason our way to the correct answer.

√x + √y > √(x+y)

To illustrate:
√4 + √9 > √(4+9)
5 > √13.

Looking at the given expression, we can quickly deduce that it must be greater than 14:
√(7 + √48) + √(7 – √48) > √(7 + √48 + 7 - √48)
√(7 + √48) + √(7 – √48) > √14

Squaring both sides:
[√(7 + √48) + √(7 – √48)]² > (√14)²
[√(7 + √48) + √(7 – √48)]² > 14

Only one answer choice is greater than 14.

The correct answer is E.
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