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If x, y, P, and Q are positive integers such that x + y = P and x2 = 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 sessions

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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. \(3P(√(P−2Q)) - 8Q\)

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D
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If x, y, P, and Q are positive integers such that x + y = P and x2 = 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
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If x, y, P, and Q are positive integers such that x + y = P and x2 = 26 Jan 2020, 10:59
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lnm87
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. \(3P(√(P−2Q)) - 8Q\)
Show more


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
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If x, y, P, and Q are positive integers such that x + y = P and x2 = 17 Jul 2020, 11:00
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sujoykrdatta
lnm87
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. \(3P(√(P−2Q)) - 8Q\)


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
Show more
Not sure what I am doing wrong. Tried 2 different sets of values for x and y, could not get D as the answer.
Set 1:
x=2, y=3, P=5, Q=6
(x-3y)(y+3x) =(2-9)(3+6) = -63
D. 3P(√(P^2−4Q)) - 8Q = 15(√25-24) - 48 = 15-48 =-33

Set 2:
x=1, y=2, P=3, Q=2
(x-3y)(y+3x) =(1-6)(2+3) = -25
D. 3P(√(P^2−4Q)) - 8Q = 9(√9-8) - 16 = 9-16 =-7

Please help, where am I going wrong?

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