Question

Ello....

Prove that

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

$\sf\ \alpha \beta= \dfrac{c}{a}$


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

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

​

Answer 1

To prove :-

• $\sf\ \alpha+ \beta= \dfrac{-b}{a}$

• $\sf\ \alpha \beta= \dfrac{c}{a}$

Proof :-

Given,

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

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

______

Now,

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

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

$:\implies\sf\ \alpha+\beta= \cancel{\dfrac{-2b}{2a}}$

$:\implies\boxed{\purple{\sf \alpha+\beta= \dfrac{-b}{a}}}$

_______

Again,

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

$:\implies\sf\ \alpha \beta= \dfrac{(-b)^2-(\sqrt{b^2-4ac})^2}{2a\times 2a}$

$:\implies\sf\ \alpha \beta= \dfrac{b^2- (b^2-4ac)}{4a^2}$

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

$:\implies\sf\ \alpha \beta= \cancel{\dfrac{4ac}{4a^2}}\\ \\ \\ :\implies\boxed{\purple{\sf \alpha \beta= \dfrac{c}{a}}}$

$\sf (Hence\ Proved....... !)$

_______________________________________________________________

Answer 2

$\large{\underline{\underline{\red{\tt{\purple{\leadsto } GiveN:-}}}}}$

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

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

$\large{\underline{\underline{\red{\tt{\purple{\leadsto } To\:Prove :-}}}}}$

• $\sf\ \alpha+ \beta= \dfrac{-b}{a}$

• $\sf\ \alpha \beta= \dfrac{c}{a}$

$\large{\underline{\underline{\red{\tt{\purple{\leadsto } ProoF:-}}}}}$

$\underline{\blue{\sf \mathscr{A}ccording\:\:\mathscr{T}o\:\mathscr{Q} uestion:-}}$

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

$\qquad\bullet\green{\sf\ \beta = \dfrac{-b-\sqrt{b^2-4ac}}{2a}}$

$\tt:\implies \alpha + \beta =\bigg\lgroup \dfrac{-b+\sqrt{b^2-4ac}}{2a}\bigg\rgroup+ \bigg\lgroup\dfrac{-b-\sqrt{b^2-4ac}}{2a} \bigg\rgroup$

$\tt:\implies\alpha+\beta =\dfrac{ -b+ \cancel{\sqrt{b^2-4ac}} - b - \cancel{\sqrt{b^2-4ac}}}{2a}$

$\tt:\implies\alpha + \beta = \dfrac{-b-b}{2a}$

$\tt:\implies \alpha+\beta=\cancel{\dfrac{-2b}{2a}}$

$\orange{\underset{\purple{\bf \: Hence\:Proved}}{\underbrace{\underline{\boxed{\red{\tt\longmapsto \alpha+\beta=\dfrac{-b}{a}}}}}}}$

$\rule{200}2$

$\tt:\implies \alpha\beta =\bigg\lgroup \dfrac{-b+\sqrt{b^2-4ac}}{2a}\bigg\rgroup \times \bigg\lgroup \dfrac{-b-\sqrt{b^2-4ac}}{2a} \bigg\rgroup$

$\tt:\implies \alpha\beta= \dfrac{ (-b-\sqrt{b^2-4ac})(-b +\sqrt{b^2-4ac})}{2a\times 3a }$

$\tt:\implies \alpha\beta =\dfrac{(-b)^2-(\sqrt{b^2-4ac})^2}{4a^2}$

$\tt:\implies \alpha\beta = \dfrac{b^2-(b^2-4ac)}{4a^2}$

$\tt:\implies \alpha\beta = \dfrac{b^2-b^2+4ac}{4a^2}$

$\tt:\implies \alpha\beta =\cancel{\dfrac{4ac}{4a^2}}$

$\orange{\underset{\purple{\bf \: Hence\:Proved}}{\underbrace{\underline{\boxed{\red{\tt\longmapsto \alpha\beta=\dfrac{c}{a}}}}}}}$

$\rule{200}2$