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Ratio of 2 solutions - GMAT Exam pack 2

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by prata » Wed Aug 03, 2016 6:19 am
Let a, b, c, and d be non-zero 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?

(1) -ab/cd
(2) -ac/bd
(3) -ad/bc
(4) ab/cd
(5) ad/bc
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by GMATGuruNY » Wed Aug 03, 2016 6:28 am
prata wrote:Let a, b, c, and d be non-zero 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?

(1) -ab/cd
(2) -ac/bd
(3) -ad/bc
(4) ab/cd
(5) ad/bc
ax(cx + d) = -b(cx+d)

ax(cx+d) + b(cx+d) = 0.

(cx+d)(ax+b) = 0.

The resulting equation is valid only if cx+d = 0 or ax+b = 0.

Root one:
cx+d = 0
cx = -d
x = -(d/c).

Root two:
ax+b = 0
ax = -b
x = -(b/a)

(root one)/(root two):
-(d/c)/-(b/a) = (d/c) * (a/b) = ad/bc.

The correct answer is E.
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by Scott@TargetTestPrep » Wed Aug 03, 2016 8:57 am
Let a, b, c, and d be non-zero 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?

(1) -ab/cd
(2) -ac/bd
(3) -ad/bc
(4) ab/cd
(5) ad/bc
ax(cx + d) = -b(cx + d)

ax(cx + d) + b(cx + d) = 0

Next we can factor out (cx + d):

(cx + d)(ax + b) = 0

Using the zero product property, we can set the expression in each set of parentheses to zero:

cx + d = 0

cx = -d

x = -d/c

Or

ax + b = 0

ax = -b

x = -b/a

So a possible ratio is:

(-d/c)/(-b/a)

ad/bc

Answer:E

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by Matt@VeritasPrep » Thu Aug 04, 2016 8:24 pm
I think I've got a quicker way.

Suppose we have

ax(cx + d) = -b(cx+d)

If (cx + d) ≠ 0, then we can divide both sides by (cx + d), leaving us

ax = -b

x = -b/a

But if (cx + d) does = 0, then we have cx + d = 0, or x = -d/c.

That gives us our two solutions, the ratio for which is

(-b/a) / (-d/c)

or

bc/ad

This is the same as E, so we're set.
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