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Let a,b,c, and d be nonzero real numbers. If the quadratic equation [#permalink]
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26 Apr 2016, 17:39
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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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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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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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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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05 May 2016, 02:58
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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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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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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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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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. If the quadratic equation [#permalink]
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03 Aug 2017, 14:29
This is a 'could be' question  don't overly complicate it.
\(ax(cx+d)=b(cx+d) \rightarrow acx^2 +adx=bcxbd \rightarrow acx^2+x(ad+bc)+bd\)
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 wellformed 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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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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28 Dec 2017, 13:36
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)




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