Question

If $\alpha ,\beta ,\gamma$ are the roots of the quadratic equation $2x^3-x^2+x-1=0$ then ${\alpha }^2+{\beta }^2+{\gamma }^2$=

• A. $\frac{5}{4}$
• B. $\frac{3}{4}$
• C. $\frac{-5}{4}$
• D. $\frac{-3}{4}$

Answer 1

The correct option is A $\frac{5}{4}$

$2x^3-x^2+x-1$

Here, $a=2,b=-1,c=1,d=-1$

$\alpha ,\beta ,\gamma$ are the roots of the above equation.

Therefore,

Sum of roots $=\frac{-b}{a}$

$\Rightarrow \alpha +\beta +\gamma =\frac{-(-1)}{2}=\frac{1}{2}$

Sum of product of roots $=\frac{c}{a}$

$\Rightarrow \alpha \beta +\beta \gamma +\gamma \alpha =\frac{1}{2}$

Product of roots $=\frac{-d}{a}$

$\alpha \beta \gamma =\frac{-(-1)}{2}=\frac{1}{2}$

Now,

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

$\Rightarrow {\alpha }^2+{\beta }^2+{\gamma }^2={\left(\frac{1}{2}\right)}^2-2\left(\frac{1}{2}\right)$

$\Rightarrow {\alpha }^2+{\beta }^2+{\gamma }^2=\frac{1}{4}-1=\frac{-3}{4}$