The value of p for which one root of the quadratic equation (p2−5p+3)x2+(3p−1)x+2=0 is twice as large as the other is
A
−23
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B
13
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C
−13
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D
23
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Solution
The correct option is D23 Let y and 2y be the roots of the quadratic equation (p2−5p+3)x2+(3p−1)x+2=0 Sum of roots: y+2y=−3p+1p2−5p+3⇒y=−3p+13(p2−5p+3)....(1) Product of roots: 2y2=2p2−5p+3....(2) Substituting (1) in (2), we get (−3p+13(p2−5p+3))2=1p2−5p+3 ⇒9p2−6p+1=9p2−45p+27 ⇒p=23 Ans: D