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Let a,b,c, and d be nonzero real numbers. If the quadratic equation

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Let a,b,c, and d be nonzero real numbers. If the quadratic equation  [#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. \(\frac{ad}{bc}\)
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Re: Let a,b,c, and d be nonzero real numbers. If the quadratic equation  [#permalink]

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New post 26 Apr 2016, 21:19
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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. If the quadratic equation  [#permalink]

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New post 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?

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

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. If the quadratic equation  [#permalink]

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New post 26 Apr 2016, 20:37
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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. If the quadratic equation  [#permalink]

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New post 02 May 2016, 19:40
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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. If the quadratic equation  [#permalink]

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New post 05 May 2016, 02:58
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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. If the quadratic equation  [#permalink]

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New post 03 Aug 2016, 10:02
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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. If the quadratic equation  [#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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Re: Let a,b,c, and d be nonzero real numbers. If the quadratic equation  [#permalink]

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New post 06 Aug 2016, 05:32
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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!

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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. If the quadratic equation  [#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. If the quadratic equation  [#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. If the quadratic equation  [#permalink]

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New post 28 Dec 2017, 13:36
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ax(cx+d)=-b(cx+d)

either cx+d = 0:
then, x = -d/c ..... (solution 1)

or cx + d not equal to 0:
then cancel it from both sides and;
ax = -b
i.e. x = -b/a ........ (solution 2)

therefore; ratio of 2 solutions = (-d/c) / (-b/a)
i.e. ratio = ad/bc ..... (answer)
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Re: Let a,b,c, and d be nonzero real numbers. If the quadratic equation  [#permalink]

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New post 13 May 2018, 03:37
VeritasPrepKarishma wrote:
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)


Excellent explanation!!!! I got to the point of (cx+d)(ax + b) = 0 but then started to multiply everything and ended up in a huge mess... Obviously that either cx+d=0 or ax+b=0.... that's a basic quadratic expression rule..
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Re: Let a,b,c, and d be nonzero real numbers. If the quadratic equation  [#permalink]

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New post 26 May 2018, 18:23
ScottTargetTestPrep wrote:
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


i don't mean to bump an old post, but i follow along your process all the way until the end where you get the ratio of ad/bc from (-d/c)/(-b/a). i'm not following. is there something basic i'm missing on why you're taking the denominator of one/numerator of the other, vis verse, and making them both positive?
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Re: Let a,b,c, and d be nonzero real numbers. If the quadratic equation  [#permalink]

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New post 01 Jun 2018, 08:39
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Rephrasing statement of question

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

Roots are -b/a or -d/c

So ratio would be bc/ad or ad/bc

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Re: Let a,b,c, and d be nonzero real numbers. If the quadratic equation  [#permalink]

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New post 04 Jan 2019, 09:10
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. \(\frac{ad}{bc}\)


The trick to answer this question correctly was to note that the equation is already written in the standard solution form. One can solve for the roots and take the fraction to get the correct answer.
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Re: Let a,b,c, and d be nonzero real numbers. If the quadratic equation  [#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. \(\frac{ad}{bc}\)


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,
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Re: Let a,b,c, and d be nonzero real numbers. If the quadratic equation  [#permalink]

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New post 12 Sep 2019, 08:25
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. \(\frac{ad}{bc}\)


Given: Let a,b,c, and d be nonzero real numbers.

Asked: 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?

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

x = - b/a or -d/c

Ratio of solutions = bc/ad or ad/bc

IMO E
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Re: Let a,b,c, and d be nonzero real numbers. If the quadratic equation   [#permalink] 12 Sep 2019, 08:25
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