Question

If $\alpha ,\beta$ are the roots of $a{x}^2+bx+c=0$, find the value of

i)$(1+\alpha )(1+\beta )$

ii)${\alpha }^3\beta +\alpha {\beta }^3$

iii)$\frac{1}{\alpha }+\frac{1}{\beta }$

iv)$\frac{1}{a\alpha +b}+\frac{1}{a\beta +b}$

Answer 1

If $\alpha ,\beta$ are roots of equation $a{x}^2+bx+c=0$ then $\alpha +\beta =\frac{-b}{a}\&\alpha \beta =\frac{c}{a}$

i) $(1+\alpha )(1+\beta )$

$=1+\alpha +\beta +\alpha \beta$

$=1+\left(\frac{-b}{a}\right)+\frac{c}{a}$

$=1+\left(\frac{c-b}{a}\right)$

ii) ${\alpha }^3\beta +\alpha {\beta }^3$

$\alpha \beta ({\alpha }^2+{\beta }^2)$

$\frac{c}{a}(\frac{b^2}{a^2}-\frac{2c}{a})$

$\frac{c}{a^3}(b^2-2ac)$

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

$(\alpha +\beta {)}^2=b^2/a^2$

${\alpha }^2+{\beta }^2+2\alpha \beta =b^2/a^2$

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

iii) $\frac{1}{\alpha }+\frac{1}{\beta }$

$\frac{\beta }{\alpha \beta }+\frac{\alpha }{\alpha \beta }=\frac{\beta +\alpha }{\alpha \beta }$

$=\frac{-b}{a}\times \frac{a}{c}=\frac{-b}{c}$

iv) $\frac{1}{a\alpha +b}+\frac{1}{a\beta +b}$ $\alpha +\beta =\frac{-b}{a}$

$=\frac{1}{a\alpha -a(\alpha +\beta )}+\frac{1}{a\beta -a(\alpha +\beta )}\Rightarrow b=-a(\alpha +\beta )$

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

$=-\frac{1}{a}(\frac{1}{\alpha }+\frac{1}{\beta })=\frac{-1}{a}\left(\frac{-b}{c}\right)=\frac{b}{ac}$