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Question

If −2 and 3 are the zeros of the quadratic polynomial x2 + (a + 1) x + b, then

(a) a = −2, b = 6
(b) a = 2, b = −6
(c) a = −2, b = −6
(d) a = 2, b = 6

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Solution

(c) a=-2, b=-6 Given: 2 and 3 are the zeroes of x2+(a+1)x+b. Now, (2)2+(a+1)×(2)+b=0=>42a2+b=0=>b2a=2 ...(1)Also, 32+(a+1)×3+b=0=>9+3a+3+b=0 =>b+3a=12 ...(2)On subtracting (1) from (2), we get a=2b=24=6 [From (1)]

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