Question

Solve above question​

Answer 1

Question 23 ) Solution Given Below⬇

$f(x) = ax {}^{2} + bx + c$

:⟼ Find

$\frac{1}{ \alpha } - \frac{1}{ \beta } = {?}^{}$

______________________________________

Required Solution ⬇

$\alpha , \beta \: are \: the \: roots \: of \: f(x)$

${\bold{\green{\underline{\underline{We \: Know \: that...}}}}}$

$\;\large{\boxed{\bf{\pink{ \alpha + \beta = \frac{ - b}{a} }}}} \: for \: f(x)$

$\;\large{\boxed{\bf{\pink{ \alpha \beta = \frac{c}{a} }}}} \: for \: f(x)$

$\frac{1}{ \alpha } - \frac{1}{ \beta } \\ ⇒ \frac{( \beta - \alpha )}{ \alpha \beta }$

$\sf\purple{ \: ( \beta - \alpha ) = {}^{} {?}^{} }$

$( \beta - \alpha ) {}^{2} = ( \beta + \alpha ) {}^{2} - 4 \beta \alpha \\ ⇒( \frac{ - b}{a} ) {}^{2} - 4( \frac{c}{a} ) \\ ⇒( \beta - \alpha ) {}^{2} = \frac{b {}^{2} }{a {}^{2} } - \frac{4c}{a} \\ \\⇒ ( \beta - \alpha ) { }^{2} = \frac{b {}^{2} - 4ac }{a {}^{2} } \\ ⇒( \beta - \alpha ) = \frac{ \sqrt{b {}^{2} - 4ac } }{a {}^{2} } \\ ( \beta - \alpha ) = \;\large{\boxed{\bf{\red{\frac{ \sqrt{b {}^{2} - 4ac } }{a {}^{2} }}}}} - - (i)$

_______________________________________

$\frac{1}{ \alpha } - \frac{1}{ \beta } = \frac{ \beta - \alpha }{ \alpha \beta }$

${\bold{\pink{\underline{\underline{using \: eq(i) \: we \: get \: ....}}}}}$

$⇒ \frac{ \frac{ \sqrt{b {}^{2} - 4ac } }{a} }{ \frac{c}{a} }$

$⇒ \frac{( \sqrt{b {}^{2} - 4ac) \times \sf\red{a} } }{ \sf\red{a}\times c}$

$\sf\colorbox{red}{required \: answer }$

$\;\large{\boxed{\bf{\green{ \frac{1}{ \alpha } - \frac{1}{ \beta } = \frac{ \sqrt{b { }^{2} - 4ac} }{c} }}}}$

$\sf\blue{hope \: this \: helps \: you!! \: }$

@Mahor2111

__________◆◇◆◇◆◇◆◇_______________

Answer 2

Answer:

achha majak tha

aisa link v koi dalta h kya ...(⊙_◎)