If $α$ and $β$ are the zeros of the quadratic polynomial $f(x)\ =\ ax^2\ +\ bx\ +\ c$, then evaluate:
$\frac{1}{α}\ -\ \frac{1}{β}$.
Given:
$α$ and $β$ are the zeros of the quadratic polynomial $f(x)\ =\ ax^2\ +\ bx\ +\ c$.
To do:
We have to find the value of $\frac{1}{α}\ -\ \frac{1}{β}$.
Solution:
The given quadratic equation is $ax^2+bx+c=0$, where a, b and c are constants and $a≠0$.
Sum of the roots $= α+β = \frac{-b}{a}$.
Product of the roots $= αβ = \frac{c}{a}$.
We know that,
$\frac{1}{\alpha } -\frac{1}{\beta } =\frac{\beta -\alpha }{\alpha \beta }$
$=\frac{-( \alpha -\beta )}{\alpha \beta }$
$( \alpha -\beta )^{2} =( \alpha +\beta )^{2} -4\alpha \beta$
$=\left( -\frac{b}{a}\right)^{2} -4\left(\frac{c}{a}\right)$
$=\frac{b^{2}}{a^{2}} -\frac{4c}{a}$
$=\frac{b^{2} -4ac}{a^{2}}$
$( \alpha -\beta ) =\sqrt{\frac{b^{2} -4ac}{a^{2}}}$
$=\frac{\sqrt{b^{2} -4ac}}{a}$
$\frac{1}{\alpha } -\frac{1}{\beta } =\frac{-( \alpha -\beta )}{\alpha \beta }$
$=-\frac{\frac{\sqrt{b^{2} -4ac}}{a}}{\frac{c}{a}}$
$=-\frac{\sqrt{b^{2} -4ac}}{a} \times \frac{a}{c}$
$=-\frac{\sqrt{b^{2} -4ac}}{c}$
The value of $\frac{1}{α}\ -\ \frac{1}{β}$ is $-\frac{\sqrt{b^{2} -4ac}}{c}$.
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