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Dear student

Since α and β are the roots of {ax}^{2} + bx + c = 0 So, α + β = - \frac{Coeff . of x}{Coeff . of x^{2}} = - \frac{b}{a} and αβ = \frac{Constant term}{Coeff . of x^{2}} = \frac{c}{a} . . . . (1) Since α_{1} and - β are the roots of a_{1} x^{2} + b_{1} x + c_{1} = 0 So, α_{1} - β = - \frac{Coeff . of x}{Coeff . of x^{2}} = - \frac{b_{1}}{a_{1}} and - α_{1} β = \frac{Constant term}{Coeff . of x^{2}} = \frac{c_{1}}{a_{1}} . . . (2) Let S and P be the sum and the product of the zeros of the required polynomial respectively . S = α + α_{1} = α + β + α_{1} - β = - \frac{b}{a} - \frac{b_{1}}{a_{1}} = - (\frac{b}{a} + \frac{b_{1}}{a_{1}}) . . . . (3) Subtract (2) from (1), we get β (α + α_{1}) = \frac{c}{a} - \frac{c_{1}}{a_{1}} - β (\frac{b}{a} + \frac{b_{1}}{a_{1}}) = \frac{c}{a} - \frac{c_{1}}{a_{1}} [using 3] - β = \frac{{ca}_{1} - {ac}_{1}}{{ba}_{1} + {ab}_{1}} β = \frac{{ac}_{1} - {ca}_{1}}{{ba}_{1} + {ab}_{1}} and P = {αα}_{1} = \frac{αβ \times α_{1} β}{β^{2}} = \frac{\frac{c}{a} \times (- \frac{c_{1}}{a_{1}})}{{(\frac{{ac}_{1} - {ca}_{1}}{{ba}_{1} + {ab}_{1}})}^{2}} = \frac{- {cc}_{1}}{{aa}_{1}} \times {(\frac{{ba}_{1} + {ab}_{1}}{{ac}_{1} - {ca}_{1}})}^{2} So, Required polynomial is x^{2} - Sx + P = x^{2} + (\frac{b}{a} + \frac{b_{1}}{a_{1}}) x - \frac{{cc}_{1}}{{aa}_{1}} \times {(\frac{{ba}_{1} + {ab}_{1}}{{ac}_{1} - {ca}_{1}})}^{2}

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