Question

if alpha,beta are the equations of axsquare+bx+c=0,then the whole root of(-b/a)whole square - 4c/a​

Answer 1

Step-by-step explanation:

$a {x}^{2} + bx + c = 0$

Dividing a on both sides

$\frac{a {x}^{2} }{a} + \frac{b}{a} x + \frac{c}{a} = 0$

${x}^{2} + \frac{b}{a} x + \frac{c}{a} = 0$

${x}^{2} + \frac{b}{a} x = - \frac{c}{a}$

Making Formula Like

(g+h)²=g²+2gh+h²

${(x)}^{2} + 2.x. \frac{b}{2a} + {( \frac{b}{2a} )}^{2} = { (\frac{b}{2a}) }^{2} - \frac{c}{a}$

Finally Formula has been made

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

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

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

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

The Equation has two root=

$\alpha \: and \: \beta \: according \: to \: ( + \: or - )$

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

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

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