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17 Jan 2018, 01:07
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
25%
(medium)
Question Stats:
83% (02:39) correct
17%
(02:29)
wrong
based on 241
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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?
\(A. 1, 8\)
\(B. -1,-8\)
\(C. 2,4\)
\(D. -1,-5\)
\(E. 1,-5\)
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?
\(A. 1, 8\)
\(B. -1,-8\)
\(C. 2,4\)
\(D. -1,-5\)
\(E. 1,-5\)
17 Jan 2018, 01:29
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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?
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?
17 Jan 2018, 11:15
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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
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
