Question

do it kal exam hai....​

Answer 1

Given :-

$\sf\ \alpha = \dfrac{-b+\sqrt{b^2-4ac}}{2a}\\ \\ \\ \sf\ \beta = \dfrac{-b-\sqrt{b^2-4ac}}{2a}$

To prove :-

$\sf\ \alpha+ \beta= \dfrac{-b}{a}\\ \\ \\ \sf\ \alpha \beta= \dfrac{c}{a}$

Proof :-

$\implies\sf\ \alpha+ \beta= \Bigg\{\dfrac{-b+\sqrt{b^2-4ac}}{2a}\Bigg\}+\Bigg\{\dfrac{-b-\sqrt{b^2-4ac}}{2a}\Bigg\}\\ \\ \\ \\ \implies\sf\ \alpha+\beta= \dfrac{-b+\cancel{\sqrt{b^2-4ac}}+(-b)-\cancel{\sqrt{b^2-4ac}}}{2a}\\ \\ \\ \\ \implies\sf\ \alpha+\beta= \cancel{\dfrac{-2b}{2a}}\\ \\ \\ \\ \implies\boxed{\purple{\sf \alpha+\beta= \dfrac{-b}{a}}}\ \ \ \ \sf( Hence\ Proved !!)$

Now,

$\implies\sf\ \alpha \beta= \Bigg\{\dfrac{-b+\sqrt{b^2-4ac}}{2a}\Bigg\}\Bigg\{\dfrac{-b-\sqrt{b^2-4ac}}{2a}\Bigg\}\\ \\ \\ \\ \implies\sf\ \alpha \beta= \dfrac{(-b)^2-(\sqrt{b^2-4ac})^2}{2a\times 2a}\\ \\ \\ \\ \implies\sf\ \alpha \beta= \dfrac{b^2- (b^2-4ac)}{4a^2}\\ \\ \\ \\ \implies\sf\ \alpha \beta= \dfrac{\cancel{b^2}-\cancel{b^2}+4ac}{4a^2}\\ \\ \\ \\ \implies\sf\ \alpha \beta= \cancel{\dfrac{4ac}{4a^2}}\\ \\ \\ \implies\boxed{\red{\sf \alpha \beta= \dfrac{c}{a}}}\ \ \ \ \sf (Hence\ Proved !!)$

Answer 2

Answer:

Given :-

$\sf\ \alpha = \dfrac{-b+\sqrt{b^2-4ac}}{2a}\\ \\ \\ \sf\ \beta = \dfrac{-b-\sqrt{b^2-4ac}}{2a}$

To prove :-

$\sf\ \alpha+ \beta= \dfrac{-b}{a}\\ \\ \\ \sf\ \alpha \beta= \dfrac{c}{a}$

Proof :-

$\implies\sf\ \alpha+ \beta= \Bigg\{\dfrac{-b+\sqrt{b^2-4ac}}{2a}\Bigg\}+\Bigg\{\dfrac{-b-\sqrt{b^2-4ac}}{2a}\Bigg\}\\ \\ \\ \\ \implies\sf\ \alpha+\beta= \dfrac{-b+\cancel{\sqrt{b^2-4ac}}+(-b)-\cancel{\sqrt{b^2-4ac}}}{2a}\\ \\ \\ \\ \implies\sf\ \alpha+\beta= \cancel{\dfrac{-2b}{2a}}\\ \\ \\ \\ \implies\boxed{\purple{\sf \alpha+\beta= \dfrac{-b}{a}}}\ \ \ \ \sf( Hence\ Proved !!)$

Now,

$\implies\sf\ \alpha \beta= \Bigg\{\dfrac{-b+\sqrt{b^2-4ac}}{2a}\Bigg\}\Bigg\{\dfrac{-b-\sqrt{b^2-4ac}}{2a}\Bigg\}\\ \\ \\ \\ \implies\sf\ \alpha \beta= \dfrac{(-b)^2-(\sqrt{b^2-4ac})^2}{2a\times 2a}\\ \\ \\ \\ \implies\sf\ \alpha \beta= \dfrac{b^2- (b^2-4ac)}{4a^2}\\ \\ \\ \\ \implies\sf\ \alpha \beta= \dfrac{\cancel{b^2}-\cancel{b^2}+4ac}{4a^2}\\ \\ \\ \\ \implies\sf\ \alpha \beta= \cancel{\dfrac{4ac}{4a^2}}\\ \\ \\ \implies\boxed{\red{\sf \alpha \beta= \dfrac{c}{a}}}\ \ \ \ \sf (Hence\ Proved !!)$