Question

If $\alpha$ is the root of the equation $x^2-x+2=0$ then the value of $\frac{6\left(-{\alpha }^3+2{\alpha }^2-\alpha \right)}{{\alpha }^5-3{\alpha }^4+3{\alpha }^3-{\alpha }^2}$ is equal to

• A. 3
• B. 6
• C. 9
• D. 12

Answer 1

Answer: (A)

${\alpha }^2-\alpha +2=0$
$\text{ So, }\frac{6\left(-{\alpha }^3+2{\alpha }^2-\alpha \right)}{{\alpha }^5-3{\alpha }^4+3{\alpha }^3-{\alpha }^2}=\frac{-6\alpha \left({\alpha }^2-2\alpha +1\right)}{{\alpha }^2\left({\alpha }^3-3{\alpha }^2+3\alpha -1\right)}$
$=\frac{-6\alpha (\alpha -1{)}^2}{{\alpha }^2(\alpha -1{)}^3}=\frac{-6}{\alpha (\alpha -1)}=\frac{-6}{{\alpha }^2-\alpha }=\frac{-6}{-2}=3$ From equation (1)