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# Two students were asked to solve the equation

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Math Revolution GMAT Instructor
Joined: 16 Aug 2015
Posts: 6834
GMAT 1: 760 Q51 V42
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Two students were asked to solve the equation  [#permalink]

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17 Jan 2018, 00:07
1
1
00:00

Difficulty:

45% (medium)

Question Stats:

90% (02:32) correct 10% (01:18) wrong based on 61 sessions

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[GMAT math practice question]

Two students were asked to solve the equation $$x^2+px+q=0$$. Alice’s answer was $$x=1$$ and $$5$$. This answer is incorrect for the original equation, but correct for the equation $$x^2+px+r=0.$$ Bob’s answer was $$x=-2$$ and $$-4$$. This answer is incorrect for the original equation, but correct for $$x^2+sx+q=0$$. What are the solutions of the original equation?

$$A. 1, 8$$
$$B. -1,-8$$
$$C. 2,4$$
$$D. -1,-5$$
$$E. 1,-5$$

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"Only $149 for 3 month Online Course" "Free Resources-30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons - try it yourself" Senior Manager Joined: 17 Oct 2016 Posts: 318 Location: India Concentration: Operations, Strategy GPA: 3.73 WE: Design (Real Estate) Re: Two students were asked to solve the equation [#permalink] ### Show Tags 17 Jan 2018, 00:29 1 sum of roots of x^2+px+r=0 is -p=1+5 (Alice's eqn) p=-6 product of roots of x^2+sx+q=0 is q=-2*(-4) (Bob's eqn) q=8 now the original equation is (x^2)-6x+8=0 whose roots are 2 and 4 which is option C But OA is B. Can someone explain? _________________ Help with kudos if u found the post useful. Thanks Intern Joined: 15 Jan 2018 Posts: 1 Re: Two students were asked to solve the equation [#permalink] ### Show Tags 17 Jan 2018, 10:15 I also think it is Answer C: x^2+px+r with x1=1 , x2=5 correspons to (x-1)*(x-5) = x^2-6x+5 x^2+sx+q with x1=-2 , x2=-4 correspons to (x+2)*(x+4) = x^2+6x+8 so x^2+px+q = x^2-6x+8 = 0 if you use pq-formula you will get x1=2 , x2=4 Manager Joined: 22 May 2017 Posts: 118 Re: Two students were asked to solve the equation [#permalink] ### Show Tags 17 Jan 2018, 10:25 I think there is some error in OA solved it individually as well (x-1)(x-5) = x2-6x+5 (x+2)(x+4) = x2+6x+8 therefore , x2-6x+8 correct equation roots are (x-2)(x-4) _________________ Kudos please if explanation helped ------------------------------------------------------------------------------------------------- Don't stop when you are tired , stop when you are DONE . Math Revolution GMAT Instructor Joined: 16 Aug 2015 Posts: 6834 GMAT 1: 760 Q51 V42 GPA: 3.82 Two students were asked to solve the equation [#permalink] ### Show Tags Updated on: 21 Jan 2018, 08:41 => Using the factor theorem with Alice’s solution yields $$(x-1)(x-5) = x^2 – 6x + 5 = x^2+px+r.$$ So, $$p = -6$$. Using the factor theorem with Bob’s solution yields $$(x+2)(x+4) = x^2 + 6x + 8 = x^2 +sx +q.$$ So, $$q = 8.$$ Thus, the original equation is $$x^2-6x+8 = (x-2)(x-4) = 0$$. Its roots are $$2$$ and $$4$$. Therefore, the answer is C. _________________ MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only$149 for 3 month Online Course"
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Originally posted by MathRevolution on 19 Jan 2018, 00:19.
Last edited by MathRevolution on 21 Jan 2018, 08:41, edited 1 time in total.
Senior Manager
Joined: 17 Oct 2016
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Location: India
Concentration: Operations, Strategy
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WE: Design (Real Estate)
Re: Two students were asked to solve the equation  [#permalink]

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19 Jan 2018, 02:03
MathRevolution wrote:
=>

Using the factor theorem with Alice’s solution yields
$$(x-1)(x-5) = x^2 – 6x + 5 = x^2+px+r.$$
So, $$p = -6$$.

Using the factor theorem with Bob’s solution yields
$$(x+2)(x+4) = x^2 + 6x + 8 = x^2 +sx +q.$$
So, $$q = 8.$$

Thus, the original equation is $$x^2-6x+8 = (x-2)(x-4) = 0$$. Its roots are $$2$$ and $$4$$.

Therefore, the answer is C.

Please clarify what is the OA? It has to be C. in that case you have to edit the OA
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Help with kudos if u found the post useful. Thanks

Math Expert
Joined: 02 Sep 2009
Posts: 52438
Re: Two students were asked to solve the equation  [#permalink]

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19 Jan 2018, 02:53
Sasindran wrote:
MathRevolution wrote:
=>

Using the factor theorem with Alice’s solution yields
$$(x-1)(x-5) = x^2 – 6x + 5 = x^2+px+r.$$
So, $$p = -6$$.

Using the factor theorem with Bob’s solution yields
$$(x+2)(x+4) = x^2 + 6x + 8 = x^2 +sx +q.$$
So, $$q = 8.$$

Thus, the original equation is $$x^2-6x+8 = (x-2)(x-4) = 0$$. Its roots are $$2$$ and $$4$$.

Therefore, the answer is C.

Please clarify what is the OA? It has to be C. in that case you have to edit the OA

The OA is C. Edited. Thank you.
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Re: Two students were asked to solve the equation &nbs [#permalink] 19 Jan 2018, 02:53
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# Two students were asked to solve the equation

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