Question

If $\alpha$ and $\beta$ are the zeroes of the polynomial $ax^{2} + bx + c = 0$, then what is the value of $\alpha -\beta$

Answer 1

$\large\underline{\sf{Solution-}}$

Given that,

$\red{\rm :\longmapsto\: \alpha , \: \beta \: are \: zeroes \: of \: {ax}^{2} + bx + c = 0}$

We know,

$\boxed{\red{\sf Product\ of\ the\ zeroes=\frac{Constant}{coefficient\ of\ x^{2}}}}$

$\bf\implies \: \alpha \beta = \dfrac{c}{a}$

and

$\boxed{\red{\sf Sum\ of\ the\ zeroes=\frac{-coefficient\ of\ x}{coefficient\ of\ x^{2}}}}$

$\bf\implies \: \alpha + \beta = - \: \dfrac{b}{a}$

Now, Consider

$\rm :\longmapsto\: \alpha - \beta$

can be rewritten as

$\rm \:  =  \: \sqrt{ {(\alpha - \beta ) \: }^{2} }$

$\rm \:  =  \: \sqrt{ {\alpha }^{2} + {\beta }^{2} - 2\alpha \beta }$

$\rm \:  =  \: \sqrt{ {\alpha }^{2} + {\beta }^{2} + 2\alpha \beta - 4\alpha \beta }$

$\rm \:  =  \: \sqrt{ {(\alpha + \beta )}^{2} - 4\alpha \beta }$

$\rm \:  =  \: \sqrt{ {\bigg[ - \dfrac{b}{a} \bigg]}^{2} - 4\bigg[\dfrac{c}{a} \bigg]}$

$\rm \:  =  \: \sqrt{\dfrac{ {b}^{2} }{ {a}^{2} } - \dfrac{4c}{a} }$

$\rm \:  =  \: \sqrt{\dfrac{ {b}^{2} - 4ac}{ {a}^{2} } }$

$\rm \:  =  \: \dfrac{ \sqrt{ {b}^{2} - 4ac} }{a}$

Hence,

$\red{ \boxed{ \sf{ \: \: \alpha - \beta =  \: \dfrac{ \sqrt{ {b}^{2} - 4ac} }{a} \: \: \: }}}$

Additional Information :-

$\red{\rm :\longmapsto\: \alpha , \beta , \gamma \: are \: zeroes \: of \: a {x}^{3} + b {x}^{2} + cx + d, \: then}$

$\boxed{ \bf{ \: \alpha + \beta + \gamma = - \dfrac{b}{a}}}$

$\boxed{ \bf{ \: \alpha\beta + \beta \gamma + \gamma \alpha = \dfrac{c}{a}}}$

$\boxed{ \bf{ \: \alpha \beta \gamma = - \dfrac{d}{a}}}$