Question

If $\alpha$ & $\beta$ are the zeroes of polynomial $x^3-12x^3+39x-28$. If it is given that the zeroes are in A.P. Find $x$

Answer 1

$x^3-12x^3+39x-28=0$

$-11x^3+39x-28=0$

$-(x-1)(11x^2+11x-28)=0$

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$11x^2+11x-28=0$

$x_{1,2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$x=\frac{-11\pm \sqrt{{11}^2-4\cdot 11(-28)}}{2\cdot 11}=\frac{-11\pm \sqrt{1353}}{22}$

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$x=1,x=\frac{-11+\sqrt{1353}}{22},x=\frac{-11-\sqrt{1353}}{22}$