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# Two students were asked to solve the equation

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Math Revolution GMAT Instructor
Joined: 16 Aug 2015
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17 Jan 2018, 01:07
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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 $99 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: 322 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, 01:29 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, 11:15 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: 113 Re: Two students were asked to solve the equation [#permalink] ### Show Tags 17 Jan 2018, 11:25 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: 6398 GMAT 1: 760 Q51 V42 GPA: 3.82 Two students were asked to solve the equation [#permalink] ### Show Tags Updated on: 21 Jan 2018, 09:41 => 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$99 for 3 month Online Course"
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Originally posted by MathRevolution on 19 Jan 2018, 01:19.
Last edited by MathRevolution on 21 Jan 2018, 09:41, edited 1 time in total.
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19 Jan 2018, 03:03
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 $$4$$.

Please clarify what is the OA? It has to be C. in that case you have to edit the OA
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19 Jan 2018, 03:53
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 $$4$$.

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, 03:53
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