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25 Jan 2020, 23:35
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D
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Difficulty:
75%
(hard)
Question Stats:
56% (02:52) correct
44%
(03:31)
wrong
based on 98
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If x, y, P, and Q are positive integers such that x + y = P and \(x^2 = \frac{Q^2}{Y^2}\), then (x-3y)(y+3x) =
A. \(3P^2 - 8Q\)
B. 3PQ-8Q
C. \(3Q^2 - 8Q\)
D. \(3P(√(P^2−4Q)) - 8Q\)
E. \(3P(√(P−2Q)) - 8Q\)
A. \(3P^2 - 8Q\)
B. 3PQ-8Q
C. \(3Q^2 - 8Q\)
D. \(3P(√(P^2−4Q)) - 8Q\)
E. \(3P(√(P−2Q)) - 8Q\)
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globaldesi
Joined: 28 Jul 2016
Last visit: 15 Jul 2026
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Location: India
Concentration: Finance, Human Resources
Schools: ISB '18 (D)
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WE:Project Management (Finance: Investment Banking)
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26 Jan 2020, 00:34
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\((x-3y)(y+3x) = 3(x^2-y^2)-8xy\)
also
\((x-y)^2 = (x+y)^2 - 4xy\)
thus
\((x-y) =\sqrt{P^2-4Q}\)
or \((x-3y)(y+3x) = 3(x^2-y^2)-8xy\)
\(=3P(\sqrt{P^2-4Q}-8Q\)
thus D
also
\((x-y)^2 = (x+y)^2 - 4xy\)
thus
\((x-y) =\sqrt{P^2-4Q}\)
or \((x-3y)(y+3x) = 3(x^2-y^2)-8xy\)
\(=3P(\sqrt{P^2-4Q}-8Q\)
thus D
General Discussion
26 Jan 2020, 10:59
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lnm87
We know that x, y, P and Q are positive, and that:
\(x + y = P\) ... (i)
\(x^2 = \frac{Q^2}{y^2}\)
\(=> xy = Q\) ... (ii)
Thus, we have:
\((x - 3y)(y + 3x)\)
\(= xy + 3x^2 - 3y^2 - 9xy\)
\(= 3(x^2 - y^2) - 8xy\)
\(= 3(x + y)(x - y) - 8xy\)
\(= 3P(x - y) - 8Q\)
Now:
\((x - y)^2 = (x + y)^2 - 4xy = P^2 - 4Q\\
=> (x - y) = \sqrt{(P^2 - 4Q)}\\
\)
Thus: \(3P(x - y) - 8Q\\
= 3P\sqrt{(P^2 - 4Q)} - 8Q\\
\)
Answer D
Alternate approach:
We know that x, y, P and Q are positive, and that:
\(x + y = P\) ... (i)
\(=> xy = Q\) ... (ii)
Let x = 8, y = 2 => P = 10 and Q = 16
Thus:
\((x - 3y)(y + 3x) = 2 * 26 = 52\)
Let us plugin P = 10 and Q = 16 in each option:
A. \(3P^2 - 8Q = 172\) - does not match
B. \(3PQ-8Q = 352\) - does not match
C. \(3Q^2 - 8Q = 640\) - does not match
D. \(3P(√(P^2−4Q)) - 8Q = 52\) - matches
E. \(3P(√(P−2Q)) - 8Q\) => cannot be computed as we have square root of a negative number - does not match
Answer D
