Question

If alpha and beta are the zeroes of the polynomial x^2+6x+9, then form a polynomial whose zeroes are -alpha and -beta.

Answer 1

$\sf\bold{\underline{Given:}}$

polynomial = $\sf{{x}^{2}+6x+9}$

Zeroes of polynomial = $\sf{\alpha\: and\:\beta}$

$\sf\bold{\underline{To \:find:}}$

Polynomial whose zeroes is $\sf{-\alpha\: and\: -\beta}$

$\sf\bold{\underline{Formula\:used:}}$

For a given polynomial $\sf{{a}^{2}+bx+c}$ whose zeroes are $\sf{\alpha \:and\: \beta}$,

$\sf{\alpha + \beta\:=\:\frac{-b}{a}}$

$\sf{\alpha \times \beta\:=\:\frac{c}{a}}$

$\sf\bold{\underline{Solution:}}$

$\sf\mapsto{For \:first\: polynomial}$

$\sf{\alpha + \beta\:=\:\frac{-6}{1}\:=\:-6-------(i)}$

$\sf{\alpha \:=\:-6-\beta-------(ii)}$

$\sf{\alpha \times \beta\:=\:\frac{9}{1}\:=\:9-------(iii)}$

$\sf\mapsto{Now\:for\:second\:polynomial}$

$\sf{(-\alpha )+ (-\beta)\:=\:\frac{-b}{a}}$

$\sf{-(-6-\beta) - \beta\:=\:\frac{-b}{a}(from\:(ii))}$

$\sf{6-{\cancel{\beta}}+{\cancel{\beta}}\:=\:\frac{-b}{a}}$

$\sf\mapsto{\frac{-b}{a}\:=\:6}$

$\sf and \sf{(-\alpha) \times (-\beta)\:=\:\frac{c}{a}=9}$

$\sf{\therefore for \:second\:polynomial}$

$\sf {k[{x}^{2}-((-\alpha) +(-\beta))x + ((-\alpha) \times (-\beta))]}$

polynomial $\sf\mapsto{k[{x}^{2}-6x+9]\:where,\: K\: is \:a \:constant}$