Question

prove that alpha+beta=-b/a​

Answer 1

Given -

$\alpha = \frac{ - b + \sqrt{ {b}^{2} - 4ac} }{2a} \: \: and \: \beta = \frac{ - b - \sqrt{ {b}^{2} - 4ac} }{2a}$

To prove

$1) \alpha + \beta = \frac{ - b}{a}$

$2) \alpha \beta = \frac{c}{a}$

Proof -

$1) = \alpha + \beta$

Now putting the values,

$= \frac{ - b + \sqrt{ {b}^{2} - 4ac} }{2a} + \frac{ - b - \sqrt{ {b}^{2} - 4ac} }{2a}$

$= \frac{ - b + \sqrt{ {b}^{2} - 4ac } + ( - b - \sqrt{ {b}^{2} - 4ac } }{2a}$

$= \frac{ - b + \sqrt{ {b}^{2} - 4ac} - b - \sqrt{ {b}^{2} - 4ac} }{2a}$

$= \frac{ - 2b}{2a}$

$= \frac{ - b}{a}$

Proved

$2) = \alpha \beta$

$= \frac{ - b + \sqrt{ {b}^{2} - 4ac } }{2a} \times \frac{ - b - \sqrt{ {b}^{2} - 4ac } }{2a}$

$= \frac{ {( - b)}^{2} - {( \sqrt{ {b}^{2} - 4ac} )}^{2} }{4a {}^{2} }$

$= \frac{ {b}^{2} - ( {b}^{2} - 4ac)}{4a {}^{2} }$

$= \frac{ {b}^{2} - {b}^{2} + 4ac }{ {4a}^{2} }$

$= \frac{4ac}{4 {a}^{2} }$

$= \frac{c}{a}$

Hence Proved

Answer 2

Step-by-step explanation:

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