Question

if alpha and beta are the roots of ax square -bx +c (a is not equal to 0 )then calculate alpha +beta​

Answer 1

$Given \: quadratic \: eqution \: ax^{2}+bx+c=0\\a≠0$

$\implies ax^{2} + bx = - c$

/* Divide each term by a ,we get */

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

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

$\implies x^{2} + 2\times x \times \frac{b}{2a} + \left( \frac{b}{2a}\right)^{2} = \left( \frac{b}{2a}\right)^{2} -\frac{c}{a}$

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

$\implies x + \frac{b}{2a} = ± \sqrt{\left(\frac{ b^{2} - 4ac}{4a^{2}}\right)}$

$\implies x = -\frac{b}{2a} ± \sqrt{\left(\frac{ b^{2} - 4ac}{4a^{2}}\right)}$

$= \frac{-b±\sqrt{b^{2} - 4ac}}{2a}$

Therefore.,

$\alpha \:and \: \beta \: are \: two \: roots \:of \\given \: quadratic \: eqution$

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

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

Therefore.,

$\pink {\alpha + \beta} \green {= \frac{-b}{a}}$

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