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655-705 Level|   Algebra|      
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KarishmaB
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Let a,b,c, and d be nonzero real numbers. If the quadratic equation 05 Nov 2020, 23:52
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rushabh27
VeritasKarishma
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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!


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

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)

VeritasKarishma Did you multiply the two equations?
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There is only one equation -

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

Taking the RHS term to LHS, we get

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

Note that if you open the brackets, you get a regular quadratic equation with x^2, x and constant terms.
But we should not open the brackets here. Because we solve the quadratic by making factors. Here, the factors are right there in front to our eyes - (cx + d) is common to both terms. Take it out.

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

So the roots are -d/c and -b/a
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Let a,b,c, and d be nonzero real numbers. If the quadratic equation 10 Mar 2021, 23:37
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Hi experts

I have seen quadratic equations question using the term 'constants'. What impact does the word 'real numbers' have on this problem? *asking out of curiosity.
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Let a,b,c, and d be nonzero real numbers. If the quadratic equation 27 Mar 2021, 07:56
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Vegita
Hi experts

I have seen quadratic equations question using the term 'constants'. What impact does the word 'real numbers' have on this problem? *asking out of curiosity.
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Hey Vegita, a real number is a number that can be represented on a number line. It just essentially means that the number is not imaginary (such as the square root of a negative number).
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Let a,b,c, and d be nonzero real numbers. If the quadratic equation 14 Mar 2022, 16:13
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KarishmaB
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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!


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

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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Hi Karishma! Would you be able to suggest similar questions to practice? Thank you
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Let a,b,c, and d be nonzero real numbers. If the quadratic equation 14 Aug 2022, 10:21
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Here's my solution in handwritten form.
Attachments

IMG-2113.jpg
IMG-2113.jpg [ 2.73 MiB | Viewed 1413 times ]

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Let a,b,c, and d be nonzero real numbers. If the quadratic equation 28 Feb 2023, 09:29
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KarishmaB
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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!


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

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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so helpful number for me
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Let a,b,c, and d be nonzero real numbers. If the quadratic equation 21 Aug 2025, 11:11
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My technique works but may not be the most time efficient.
Plug in numbers I chose a=5 b=1 c=2 d=3

I set everything equal to zero and had a weird quadratic and used quadratic formula and got -3/2 and -1/5 as solutions.

I set an equal denominator and got 15:2 as the ratio.

I scanned for which answer choice would give me the same ratio with the plugged in numbers from before and it was correct.
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