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GRE Prep Club Team Member
Joined: 20 Feb 2017
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Let a,b,c, and d be nonzero real numbers. If the quadratic equation
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30 Mar 2022, 05:04
30 Mar 2022, 05:04
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Let a,b,c, and d be nonzero real numbers. If the quadratic equation \(ax(cx+d)=-b(cx+d)\) is solved for x, which of the following is a possible ratio of the 2 solutions?
A. \(-\frac{ab}{cd}\)
B. \(-\frac{ac}{bd}\)
C. \(-\frac{ad}{bc}\)
D. \(\frac{ab}{cd}\)
E. \(\frac{ad}{bc}\)
A. \(-\frac{ab}{cd}\)
B. \(-\frac{ac}{bd}\)
C. \(-\frac{ad}{bc}\)
D. \(\frac{ab}{cd}\)
E. \(\frac{ad}{bc}\)
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Joined: 10 Apr 2015
Posts: 6218
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Re: Let a,b,c, and d be nonzero real numbers. If the quadratic equation
[#permalink]
30 Mar 2022, 06:21
30 Mar 2022, 06:21
GeminiHeat wrote:
Let a,b,c, and d be nonzero real numbers. If the quadratic equation \(ax(cx+d)=-b(cx+d)\) is solved for x, which of the following is a possible ratio of the 2 solutions?
A. \(-\frac{ab}{cd}\)
B. \(-\frac{ac}{bd}\)
C. \(-\frac{ad}{bc}\)
D. \(\frac{ab}{cd}\)
E. \(\frac{ad}{bc}\)
A. \(-\frac{ab}{cd}\)
B. \(-\frac{ac}{bd}\)
C. \(-\frac{ad}{bc}\)
D. \(\frac{ab}{cd}\)
E. \(\frac{ad}{bc}\)
GIVEN: ax(cx + d) = -b(cx + d)
Add b(cx + d) to both sides to get: ax(cx + d) + b(cx + d) = 0
Rewrite as: (ax + b)(cx + d) = 0
This means either (ax + b) = 0 or (cx + d) = 0
Let's examine each case
Case a: ax + b = 0
So: ax = -b
Solve: x = -b/a
Case b: cx + d = 0
So: cx = -d
Solve: x = -d/c
Which of the following is a possible ratio of the 2 solutions?
ASIDE: Why does the question ask for a POSSIBLE ratio?
Well, the ratio of the solutions can be EITHER (-b/a)/(-d/c) OR (-d/c)/(-b/a)
Let's check the first ratio.
(-b/a)/(-d/c) = (-b/a)(c/-d) = bc/ad check the answer choices . . . not there.
However, answer choice E is the reciprocal of bc/ad, so it must be the correct answer.
Not convinced?
Let's confirm.
(-d/c)/(-b/a) = (-d/c)(a/-b) = ad/bc = answer choice E
Cheers,
Brent
gmatclubot
Re: Let a,b,c, and d be nonzero real numbers. If the quadratic equation [#permalink]
30 Mar 2022, 06:21
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