Question

If $\alpha $ and $\beta $ are the roots of the equation $x^{2}-2x+3=0,$ then the sum of roots of the equation having roots as $\alpha ^{3}-3\alpha ^{2}+5\alpha -2$ and $\beta ^{3}-\beta ^{2}+\beta +5$ is

• A. $1$
• B. $3$
• C. $5$
• D. $7$

Answer 1

Answer: (B)

Since, $\alpha ^{2}-2\alpha +3=0$
$\Rightarrow \left(\alpha \right)^{3}-3\left(\alpha \right)^{2}+5\alpha -2=\alpha \left(\left(\alpha \right)^{2} - 2 \alpha + 3\right)-\left(\left(\alpha \right)^{2} - 2 \alpha + 3\right)+1=1$
And $\beta ^{2}-2\beta +3=0$
$\Rightarrow \left(\beta \right)^{3}-\left(\beta \right)^{2}+\beta +5=\beta \left(\left(\beta \right)^{2} - 2 \beta + 3\right)+\left(\left(\beta \right)^{2} - 2 \beta + 3\right)+2=2$
So, the required equation is
$x^{2}-\left(2 + 1\right)x+2.1=0$
$\Rightarrow x^{2}-3x+2=0$
Sum of roots $=3$