Question

If $x_1$ and $x_2$ are the roots of $3x^2-2x-6=0$, then $x_1^2+x_2^2$ is equal to

• A. $\frac{50}{9}$
• B. $\frac{40}{9}$
• C. $\frac{30}{9}$
• D. $\frac{20}{9}$
• E. $\frac{10}{9}$

Answer 1

Answer: (B)

We have,
$3x^2-2x-6=0$
Since, $x_1$ and $x_2$ are the roots of above equation.
$\therefore x_1+x_2=\frac{-(-2)}{3}=\frac{2}{3}$
and $x_1{x}_2=\frac{-6}{3}=-2$
Now,
${\left(x_1+x_2\right)}^2=x_1^2+x_2^2+2x_1{x}_2$
$\Rightarrow x_1^2+x_2^2={\left(x_1+x_2\right)}^2-2x_1{x}_2$
$={\left(\frac{2}{3}\right)}^2-2(-2)$
$=\frac{4}{9}+4=\frac{40}{9}$