Question

If $\alpha$ and $\beta$ are the roots of the equation $2x^2-3x-1=0$, find the values of.
(i) ${\alpha }^2+{\beta }^2$
(ii) $\frac{\alpha }{\beta }+\frac{\beta }{\alpha }$
(iii) $\alpha -\beta$ if $\alpha >\beta$(iv) $(\frac{{\alpha }^2}{\beta }+\frac{{\beta }^2}{\alpha })$
(v) $(\alpha +\frac{1}{\beta })(\frac{1}{\alpha }+\beta )$
(vi) ${\alpha }^4+{\beta }^4$
(vii) $\frac{{\alpha }^3}{\beta }+\frac{{\beta }^3}{\alpha }$

Answer 1

Given equation is $2x^2-3x-1=0$
Let the given equation be written as $a{x}^2+bx+c=0$
Then, a =2, b = 3, c = -1.
Given a and b are the roots of the equation.
$\therefore \alpha +\beta =\frac{-b}{a}=\frac{-(-3)}{2}=\frac{3}{2}$ and $\alpha \beta =-\frac{1}{2}$
(i) ${\alpha }^2+{\beta }^2=(\alpha +\beta {)}^2-2\alpha \beta ={\left(\frac{3}{2}\right)}^2-2(-\frac{1}{2})=\frac{9}{4}+1=\frac{13}{4}$
(ii) $\frac{\alpha }{\beta }+\frac{\beta }{\alpha }=\frac{{\alpha }^2+{\beta }^2}{\alpha \beta }=\frac{(\alpha +\beta \beta {)}^2-2\alpha \beta }{\alpha \beta }=\frac{{\left(\frac{3}{2}\right)}^2-2(-\frac{1}{2})}{-\frac{1}{2}}=\frac{13}{4}\times (-2)=-\frac{13}{2}$
(iii) $\alpha -\beta =\sqrt{(\alpha +\beta {)}^2-4\alpha \beta }$
$={[{\left(\frac{3}{2}\right)}^2-4\times (-\frac{1}{2})]}^{\frac{1}{2}}={(\frac{9}{4}+2)}^{\frac{1}{2}}=\frac{\sqrt{17}}{2}$