Question

If $\alpha$ and $\beta$ are the roots of the equation $3x^2-6x+4=0$, fid the value of

${\alpha }^2$ + ${\beta }^2$.

Answer 1

Given Equation: $3x^2-6x+4=0$
We know that, $Ifa{x}^2+bx+c=0$ is a quadratic equation then root are $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Substituting the values of $a,b$ and $c$

$x=\frac{-6\pm \sqrt{36-48}}{6}$

$\Rightarrow \alpha =\frac{-6+\sqrt{1}2i}{6}\&\beta =\frac{-6-\sqrt{1}2i}{6}$

$\Rightarrow \alpha =\frac{-3+\sqrt{3}i}{3}\&\beta =\frac{-3-\sqrt{3}i}{3}$

Finding ${\alpha }^2+{\beta }^2$

$={\left(\frac{-3+\sqrt{3}i}{3}\right)}^2+{\left(\frac{-3-\sqrt{3}i}{3}\right)}^2$

$=\frac{9-6\sqrt{3}i-3}{9}+\frac{9+6\sqrt{3}i-3}{9}$

$=\frac{12}{9}$ = $\frac{4}{3}$

$\therefore$ The value of ${\alpha }^2+{\beta }^2$ = $\frac{4}{3}$