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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, 00:07
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Difficulty:

55% (hard)

Question Stats:

82% (02:10) correct 18% (00:44) wrong based on 49 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?

$$A. 1, 8$$
$$B. -1,-8$$
$$C. 2,4$$
$$D. -1,-5$$
$$E. 1,-5$$
[Reveal] Spoiler: OA

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17 Jan 2018, 00:29
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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?
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Intern
Joined: 15 Jan 2018
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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
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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)
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19 Jan 2018, 00:19
=>

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 $$4$$.

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Last edited by MathRevolution on 21 Jan 2018, 08:41, edited 1 time in total.
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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 $$4$$.

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, 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 $$4$$.

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   [#permalink] 19 Jan 2018, 02:53
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