if alpha and beta are zeroes of x^2-x-2 form a quadratic polynomial in x whose zeroes are ( 2alpha +1) and (2beta +1)

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Solution

$The given polynomial is, p (x) = x^{2} - x - 2 Now, sum of zeroes of p (x), α + β = - \frac{coefficient of x}{coefficient of x^{2}} = - \frac{(- 1)}{1} = 1 product of zeroes of p (x), α β = \frac{constant term}{coefficient of x^{2}} = - 2$

$Since (2 α + 1) and (2 β + 1) are the zeroes of required polynomial, then sum of zeroes of required polynomial, S = (2 α + 1) + (2 β + 1) = 2 (α + β) + 2 = 2 (1) + 2 = 4 product of zeroes of required polynomial, P = (2 α + 1) (2 β + 1) = = 4 α β + 2 (α + β) + 1 = 4 (- 2) + 2 (1) + 1 = - 3 So, required polynomial, is f (x) = x^{2} - S x + P = x^{2} - 4 x - 3$

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