Question

If Sum of zeros of a polyhomial $=\alpha +\beta =-8$ and product of zeroes $=\alpha \beta =6$ then form a polynomial whose zeroes are $(\alpha -\beta )and(\alpha +\beta )$

Answer 1

Given $\alpha +\beta =-8$ & $\alpha \beta =6$

$(\alpha +\beta {)}^2={\alpha }^2+{\beta }^2+2\alpha \beta$

${\alpha }^2+{\beta }^2=64-12=52$

$(\alpha -\beta {)}^2={\alpha }^2+{\beta }^2-2\alpha \beta$

$(\alpha -\beta {)}^2=52-12=40$

$(\alpha -\beta )=2\sqrt{10}$

now $(\alpha +\beta ),(\alpha -\beta )$ are roots of any quadratic equation,

quadratic equation is

$[x-(\alpha +\beta \left)\right][x-(\alpha -\beta \left)\right]=0$

$(x+8)(x-2\sqrt{10})=0$

$x^2+(8-2\sqrt{10})x-16\sqrt{10}=0$

Hence equation (1) is required equation.