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Let a,b,c, and d be nonzero real numbers

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

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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!
[Reveal] Spoiler: OA

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

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amitpaul527 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. \(\frac{ad}{bc}\)


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--

\(\frac{S+T}{ST} = -(\frac{ad+bc}{ac})/\frac{bd}{ac}\)..
\(\frac{S}{ST}+\frac{T}{ST} = -( \frac{ad}{bd} + \frac{bc}{bd})\) ..
\(\frac{1}{T} + \frac{1}{S} = -\frac{a}{b} - \frac{c}{d}\)..

so JUST for finding the ratio, we will equate two sides--
Here it doesn't matter what do you equate with, because we are just finding any of the ratio

\(\frac{1}{T}= -\frac{a}{b}\)and \(\frac{1}{S}= -\frac{c}{d}\)..

\(\frac{\frac{1}{T}}{\frac{1}{S}}\) = \(\frac{-\frac{a}{b}}{-\frac{c}{d}}\)..

\(\frac{S}{T} = \frac{ad}{bc}\)

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

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amitpaul527 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. \(\frac{ad}{bc}\)


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

Roots are -b/a and -d/c

Their ratio could be bc/ad or ad/bc.

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

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

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dgeorgie 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 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

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

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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?

\(Let\mbox{'}s\; say\; a=1,\; b=2,\; c=3\; and\; d=4\)

the solutions would be:
a) -\(\frac{1}{6}\)
b) -\(\frac{3}{8}\)
c) -\(\frac{2}{3}\)
d) \(\frac{1}{6}\)
e) \(\frac{2}{3}\)


\(Now\; let\mbox{'}s\; solve\; the\; equation,\; after\; substituting\; a=1,\; b=2,\; c=3\; and\; d=4\; into\; the\; eq.\; we\; find:
3x^{2}\; +\; 10x\; +\; 8\; =\; 0\)
\(which\; solved\; gives\; us\; \left( x+\frac{4}{3} \right)\left( x+2 \right),\; where\; the\; roots\; are\; x_{a}=\; -\frac{4}{3}\; and\; x_{b}=\; -2,\; if\; we\; divide\; them:\; \frac{-\frac{4}{3}}{-2}\; we\; find\; \frac{xa}{xb}\; =\; \frac{2}{3}\)

\(Answer\; E\)

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

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amitpaul527 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. \(\frac{ad}{bc}\)


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

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New post 06 Aug 2016, 03:06
Hello, can someone please explain why this works:
ax(cx + d) = -b(cx + d)
axc^2 + axd = -bcx - bd
ax^2 + axd + bcx + bd = 0
ax^2 + (ad + bc)x + bd = 0

Then take the ratio of the ad/bc (in the brackets) - is it just luck that I got it right this way? Thanks for the help!

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

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New post 06 Aug 2016, 05:32
AfricanPrincess wrote:
Hello, can someone please explain why this works:
ax(cx + d) = -b(cx + d)
axc^2 + axd = -bcx - bd
ax^2 + axd + bcx + bd = 0
ax^2 + (ad + bc)x + bd = 0

Then take the ratio of the ad/bc (in the brackets) - is it just luck that I got it right this way? Thanks for the help!

Posted from my mobile device


In the highlighted portion above, you have missed c in acx^2.

Also, in the bracket you have the sum of ad and bc not the product.

Now, if I consider your statement

acx^2 + (ad + bc)x + bd = 0

There is a very good solution provided by chetan2u in the forum. You can refer to that solution.

Else, you can simply do it in this way :

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

=> either cx+d=0 => x=-d/c ---(1)
or
ax=-b => x=-b/a ---(2)

Divide (1) by (2) or vice versa, you will get ad/bc or bc/ad.
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Re: Let a,b,c, and d be nonzero real numbers [#permalink]

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New post 03 Aug 2017, 14:29
This is a 'could be' question - don't overly complicate it.

\(ax(cx+d)=-b(cx+d) \rightarrow acx^2 +adx=-bcx-bd \rightarrow acx^2+x(ad+bc)+bd\)

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

Answer E

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

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New post 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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Re: Let a,b,c, and d be nonzero real numbers   [#permalink] 18 Aug 2017, 19:23
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