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Two students were asked to solve the equation

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Two students were asked to solve the equation  [#permalink]

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New post 17 Jan 2018, 01:07
1
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A
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C
D
E

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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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Re: Two students were asked to solve the equation  [#permalink]

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New post 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?
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Re: Two students were asked to solve the equation  [#permalink]

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New post 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
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Re: Two students were asked to solve the equation  [#permalink]

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New post 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)
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Two students were asked to solve the equation  [#permalink]

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New post 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.
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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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Re: Two students were asked to solve the equation  [#permalink]

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New post 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 \(4\).

Therefore, the answer is C.
Answer: B


Please clarify what is the OA? It has to be C. in that case you have to edit the OA
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Re: Two students were asked to solve the equation  [#permalink]

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New post 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 \(4\).

Therefore, the answer is C.
Answer: B


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, 03:53
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