Question

Prove that
.

$in \: the \: equation \\ a {x}^{2} + bx + c = 0 \\ \\ \alpha + \beta = \frac{ - b}{a } \\ and \\ \alpha \beta = \frac{c}{a}$
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Answer 1

Step-by-step explanation:

$heloo... \\ \\ equation = > \\ a {x}^{2} + bx + c = 0 \\ \\ will \: have \: two \: roots = > \\ ( \alpha \: and \: \beta ) \\ \\ (x - \alpha )(x - \beta ) = 0 \\ this \: will \: be \: root \\ multiplying \: them = > \\ {x}^{2} - \alpha x - \beta x + \alpha \beta = 0 \\ {x}^{2} - ( \alpha + \beta )x + \alpha \beta = 0 \\ now \: comparing \: with \: equation \\ \frac{1}{a} = \frac{ - ( \alpha + \beta )}{b} = \frac{ \alpha \beta }{c} \\ from \: here \: we \: will \: get = > \\ \alpha + \beta = \frac{ - b}{a} \: \\ and \\ \alpha \beta = \frac{c}{a}$

Answer 2

Answer:

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