Question

# The zero of the quadratic polynomial x2+(a+1)x+b are 2 and −3, thena) a=–7,b=–1 b) a=5,b=–1 c) a=2,b=–6 c) a=0,b=–6

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Solution

## Let p(x)=x2+(a+1)x+b Given that, 2 and −3 are the zeroes of the quadratic polynomial p(x) ∴ p(2)=0 and p(−3)=0, Solving for p(2), ⇒22+(a+1)(2)+b=0 ⇒4+2a+2+b=0 ⇒2a+b=−6⋯(i) Solving for p(−3), ⇒(−3)2+(a+1)(−3)+b=0 ⇒9–3a–3+b=0 ⇒3a–b=6⋯(ii) On adding Eqs. (i) and (ii), we get 5a=0⇒a=0Substituting the value of a in Eq. (i), we get x×0+b=−6⇒b=−6So, the required values are a=0 and b=−6.

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