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28 Jan 2022, 05:15
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
(hard)
Question Stats:
68% (02:31) correct
32%
(02:41)
wrong
based on 222
sessions
History
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Not Attempted Yet
If b and c are positive integers such that \(\sqrt{11+b√c} + \sqrt{11-b√c}=6\) where \(b\sqrt{c}\) is in simplest radical form, what is the value of \(b^{2}c\)?
(A) 72
(B) 76
(C) 80
(D) 82
(E) 90
(A) 72
(B) 76
(C) 80
(D) 82
(E) 90
28 Jan 2022, 06:15
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Bunuel
\(\sqrt{11+b√c} + \sqrt{11-b√c}=6\)
Square both the sides
\((\sqrt{11+b√c} + \sqrt{11-b√c})^2=6^2\)
\(11+b\sqrt{c}+11-b\sqrt{c}+2*\sqrt{11+b√c} * \sqrt{11-b√c}=36\)
\(22+2*\sqrt{11^2-(b√c)^2}=36\)
\(\sqrt{121-b^2c}=\frac{36-22}{2}=7\)
Again square both sides
\(121-b^2c=49……..b^2c=72\)
A
28 Jan 2022, 10:31
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Bunuel
Let us replace \(b√c = a\); we need to determine the value of \(b^{2}c = a^2\)
Approach 1:
\(\sqrt{11+a} + \sqrt{11-a}=6\)
Squaring: \(11 + a + 11 - a + 2\sqrt{(11+a)(11-a)} = 36\)
\(=> 22 + 2\sqrt{(121-a^2)} = 36\)
\(=> 2\sqrt{(121-a^2)} = 14\)
\(=> \sqrt{(121-a^2)} = 7\)
Squaring: \(121-a^2 = 49\)
\(=> a^2 = 72\)
ALTERNATE:
\(\sqrt{11+a} + \sqrt{11-a}=6\)
Let P = \(\sqrt{11+a}\) and Q = \(\sqrt{11-a}\)
Thus: \(P + Q=6\) ... (i)
Thus: \(P^2 - Q^2 = (11+a) - (11-a) = 2a\) ... (ii)
Dividing (ii) by (i):
\(P - Q = a/3\) ... (iii)
Adding (i) and (iii): \(2P = 6 + a/3\)
\(=> P = (18+a)/6\)
Thus: \(P = \sqrt{11+a} = (18+a)/6\)
\(=> \sqrt{11+a} = (18+a)/6\)
Squaring: \(11 + a = (324 + a^2 + 36a)/36\)
\(=> 396 + 36a = 324 + a^2 + 36a\)
\(=> a^2 = 72\)
