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
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
Updated On: May 18, 2024
- $\frac{50}{9}$
- $\frac{40}{9}$
- $\frac{30}{9}$
- $\frac{20}{9}$
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The Correct Option is B
Solution and Explanation
We have,
$3 x^{2}-2 x-6=0$
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,
$\left(x_{1}+x_{2}\right)^{2}=x_{1}^{2}+x_{2}^{2}+2 x_{1} x_{2}$
$\Rightarrow x_{1}^{2}+x_{2}^{2}=\left(x_{1}+x_{2}\right)^{2}-2 x_{1} x_{2}$
$=\left(\frac{2}{3}\right)^{2}-2(-2)$
$=\frac{4}{9}+4=\frac{40}{9}$
$3 x^{2}-2 x-6=0$
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,
$\left(x_{1}+x_{2}\right)^{2}=x_{1}^{2}+x_{2}^{2}+2 x_{1} x_{2}$
$\Rightarrow x_{1}^{2}+x_{2}^{2}=\left(x_{1}+x_{2}\right)^{2}-2 x_{1} x_{2}$
$=\left(\frac{2}{3}\right)^{2}-2(-2)$
$=\frac{4}{9}+4=\frac{40}{9}$
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