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If zeroes of the polynomial p(x)=x^2+(a+b)x+b are 2 and -3, then find the value of (a+b) |
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Answer» <P>Answer: p(x) = x²+(a+b)x+b p = 1 q = (a+b) R = b Let α & β be the ZEROES of p(x), α = 2 β = -3 α+β = -q/p 2+(-3) = -(a+b) /1 -1 = -a+b b = a-1 -----------(1) αβ = r/p (2)(-3) = b/1 -6 = b/1 -6 = b b = -6 Substitute value of b in equation (1), -6 = a-1 -6+1 = a a = -5 So a = -5 and b = -6 a+b = -5+(-6) = -5-6 = -11 |
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