Author 
Message 
TAGS:

Hide Tags

Intern
Joined: 15 Dec 2015
Posts: 5
Concentration: Sustainability, Strategy

Let a,b,c, and d be nonzero real numbers. If the quadratic equation
[#permalink]
Show Tags
26 Apr 2016, 17:39
Question Stats:
64% (02:17) correct 36% (02:33) wrong based on 820 sessions
HideShow timer Statistics
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. \(\frac{ad}{bc}\)
Official Answer and Stats are available only to registered users. Register/ Login.




Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8373
Location: Pune, India

Re: Let a,b,c, and d be nonzero real numbers. If the quadratic equation
[#permalink]
Show Tags
26 Apr 2016, 21:19
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)
_________________
Karishma Veritas Prep GMAT Instructor
Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >
GMAT selfstudy has never been more personalized or more fun. Try ORION Free!




Manager
Status: 2 months to go
Joined: 11 Oct 2015
Posts: 116
GPA: 3.8

Re: Let a,b,c, and d be nonzero real numbers. If the quadratic equation
[#permalink]
Show Tags
26 Jul 2016, 00:47
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) \(\frac{2}{3}\)
\(Now\; let\mbox{'}s\; solve\; the\; equation,\; after\; substituting\; a=1,\; b=2,\; c=3\; and\; d=4\; into\; the\; eq.\; we\; find: 3x^{2}\; +\; 10x\; +\; 8\; =\; 0\) \(which\; solved\; gives\; us\; \left( x+\frac{4}{3} \right)\left( x+2 \right),\; where\; the\; roots\; are\; x_{a}=\; \frac{4}{3}\; and\; x_{b}=\; 2,\; if\; we\; divide\; them:\; \frac{\frac{4}{3}}{2}\; we\; find\; \frac{xa}{xb}\; =\; \frac{2}{3}\)
\(Answer\; E\)




Math Expert
Joined: 02 Aug 2009
Posts: 6957

Re: Let a,b,c, and d be nonzero real numbers. If the quadratic equation
[#permalink]
Show Tags
26 Apr 2016, 20:37
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
_________________
1) Absolute modulus : http://gmatclub.com/forum/absolutemodulusabetterunderstanding210849.html#p1622372 2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html 3) effects of arithmetic operations : https://gmatclub.com/forum/effectsofarithmeticoperationsonfractions269413.html
GMAT online Tutor



Current Student
Joined: 12 Aug 2015
Posts: 2638

Re: Let a,b,c, and d be nonzero real numbers. If the quadratic equation
[#permalink]
Show Tags
02 May 2016, 19:40



SVP
Joined: 06 Nov 2014
Posts: 1885

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



Target Test Prep Representative
Status: Founder & CEO
Affiliations: Target Test Prep
Joined: 14 Oct 2015
Posts: 3816
Location: United States (CA)

Re: Let a,b,c, and d be nonzero real numbers. If the quadratic equation
[#permalink]
Show Tags
03 Aug 2016, 10:02
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
_________________
Scott WoodburyStewart
Founder and CEO
GMAT Quant SelfStudy Course
500+ lessons 3000+ practice problems 800+ HD solutions



Intern
Joined: 20 Oct 2014
Posts: 3

Re: Let a,b,c, and d be nonzero real numbers. If the quadratic equation
[#permalink]
Show Tags
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!
Posted from my mobile device



Board of Directors
Status: Stepping into my 10 years long dream
Joined: 18 Jul 2015
Posts: 3627

Re: Let a,b,c, and d be nonzero real numbers. If the quadratic equation
[#permalink]
Show Tags
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.
_________________
My GMAT Story: From V21 to V40 My MBA Journey: My 10 years long MBA Dream My Secret Hacks: Best way to use GMATClub  Importance of an Error Log! Verbal Resources: All SC Resources at one place  All CR Resources at one place Blog: Subscribe to Question of the Day Blog
GMAT Club Inbuilt Error Log Functionality  View More. New Visa Forum  Ask all your Visa Related Questions  here.
New! Best Reply Functionality on GMAT Club! Find a bug in the new email templates and get rewarded with 2 weeks of GMATClub Tests for free



Intern
Joined: 06 Apr 2017
Posts: 29
Location: United States (OR)
Concentration: Finance, Leadership
GMAT 1: 730 Q48 V44 GMAT 2: 730 Q49 V40
GPA: 3.98
WE: Corporate Finance (Health Care)

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



Intern
Joined: 18 Jul 2017
Posts: 5

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



Intern
Joined: 16 Oct 2017
Posts: 1

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



Intern
Joined: 10 Jun 2017
Posts: 23

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



Intern
Joined: 07 May 2015
Posts: 26

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



Director
Joined: 02 Oct 2017
Posts: 645

Re: Let a,b,c, and d be nonzero real numbers. If the quadratic equation
[#permalink]
Show Tags
01 Jun 2018, 08:39
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 Give kudos if it helps Posted from my mobile device
_________________
Give kudos if you like the post




Re: Let a,b,c, and d be nonzero real numbers. If the quadratic equation &nbs
[#permalink]
01 Jun 2018, 08:39






