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01 May 2018, 01:21
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
45%
(medium)
Question Stats:
69% (01:47) correct
31%
(01:59)
wrong
based on 432
sessions
History
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Not Attempted Yet
If 3 is one root of the quadratic equation \(2x^2 - 5x + k = 2\), where k is a constant, what is the equation’s other root?
(A) -3
(B) -1/2
(C) -1
(D) 1/2
(E) 3
(A) -3
(B) -1/2
(C) -1
(D) 1/2
(E) 3
Updated on: 14 May 2018, 03:42
Originally posted by GMATinsight on 01 May 2018, 01:36.
Last edited by GMATinsight on 14 May 2018, 03:42, edited 2 times in total.
Last edited by GMATinsight on 14 May 2018, 03:42, edited 2 times in total.
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Bunuel
If 3 is one of the roots then it must satisfy the equation \(2x^2 - 5x + k = 2\)
i.e. \(2(3)^2 - 5(3) + k = 2\)
i.e. \(18 - 15 +k = 2\)
i.e. \(k = -1\)
Now the equation becomes \(2x^2 - 5x -3 = 0\)
Now we have three option
1) Either substitute options and whichever option satisfies the equation will be the correct answer
2) Sum of the roots = \(-b/a\) in a quadratic equation \(ax^2 - bx + c = 0\)
i.e. sum of the roots of \(2x^2 - 5x -3 = 0\) is given by
Sum = - (-5)/2 = 5/2
i.e. 3 + p = 5/2 . { If p is the second root and we know that 3 is the first root }
i.e. p = -1/2
3) Factorise the equation
\(2x^2 - 5x - 1 = 2\).
\(2x^2 - 5x - 3 = 0\).
\(2x^2 - 6x + x - 3 = 0\).
2x(x-3) +1(x-3) = 0
(2x+1) (x-3) = 0.
x = -1/2 and x =3 are the two roots.
One of the roots is given as 3, the other root is x=-1/2
Answer: Option B
13 May 2018, 09:10
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