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Let a,b,c, and d be nonzero real numbers [#permalink]

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26 Apr 2016, 17:39

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E

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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. \(\frac{ad}{bc}\)

Having trouble with this problem and would appreciate if someone could walk me through a solution. Thanks!

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. \(\frac{ad}{bc}\)

Having trouble with this problem and would appreciate if someone could walk me through a solution. Thanks!

Hi, one method would be take advantage of SUM and PRODUCT of roots of a quadratic equation.. \(ax(cx+d)=-b(cx+d)\) .. \(acx^2+(ad+bc)x+bd = 0\).. Let the roots be S and T Sum of roots = \(S+T = -(\frac{ad+bc}{ac})\).. Product of roots = \(S*T = \frac{bd}{ac}\)..

two methods..

1) equate from SUM.. we know \(S+T = -(\frac{ad+bc}{ac})\).. \(S+T = -(\frac{ad}{ac}+\frac{bc}{ac})\) let \(S = -\frac{ad}{ac}\)and \(T = -\frac{bc}{ac}\) so\(\frac{S}{T} = \frac{ad}{bc}\) E

2)we are to find\(\frac{S}{T} or \frac{T}{S}\).. divide SUM by PRODUCT--

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. \(\frac{ad}{bc}\)

Having trouble with this problem and would appreciate if someone could walk me through a solution. Thanks!

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

You solve a quadratic by splitting it into factors.

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

But recognise that the factors are already there in this equation. Take (cx + d) common. \((ax + b)*(cx + d) = 0\)

Re: Let a,b,c, and d be nonzero real numbers [#permalink]

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02 May 2016, 19:40

Don't Even Think about calculating the roots by opening the brackets It becomes a mess..! Here take cx+d common and solve for x we can see that the ratio may be ad/bc or bc/ad Hence E
_________________

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 ration of the 2 solutions?

A. -ab/(cd) B. -ac/(bd) C. -ad/(bc) D. ab/(cd) E. ad/(bc)

Actually, I dont understand why do we have a quadratic equation. Isn't it the same as ax=-b ? Then we would have only 1 solution I appreciate your suggestions. Thank you in advance!

No, you cannot cancel (cx+d) from both the sides. Because (cx+d) can be 0 too and you cannot cancel 0 from both sides. You can easily cancel out any numbers, but not variables.

For Example: x(x+2) = 5(x+2). In this case, you cannot simply cancel out (x+2) from both the sides.

Therefore ax(cx+d)=-b(cx+d) should be written as (ax+b)(cx+d) = 0 Now in the quadratic equation, any of the terms can be equal to zero.

Therefore, the two solutions will be x = -b/a and x = -d/c

Ratio of two roots = (-d/c) / (-b/a) = ad/bc Correct option: E

Re: Let a,b,c, and d be nonzero real numbers [#permalink]

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26 Jul 2016, 00:47

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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 ration of the 2 solutions?

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. \(\frac{ad}{bc}\)

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
_________________

Scott Woodbury-Stewart Founder and CEO

GMAT Quant Self-Study Course 500+ lessons 3000+ practice problems 800+ HD solutions

Quadratics are easier to work with when the leading coefficient is \(1\)

Since this is a COULD BE question, then let's say \(a=c=1\)

Now we have

\(x^2+x(ad+bc)+bd\)

Since this is a COULD BE question with a well-formed quadratic, the roots could definitely be \(ad/bc\), since \(ad\) and \(bc\) sum to the coefficient of the second term and multiply to the third term

Re: Let a,b,c, and d be nonzero real numbers [#permalink]

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18 Aug 2017, 19:23

one method would be take advantage of SUM and PRODUCT of roots of a quadratic equation.. acx^2+x(ad+bc)+bd=0 can anyone plz explain how to solve this equation thank you

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