Question

If $\alpha $ is a root of the equation $4x^{2}+2x-1=0$ and $f\left(x\right)=4x^{3}-3x+1,$ then then $2(f(\alpha)+(\alpha))$ is equal to

• A. $-1$
• B. $0$
• C. $1$
• D. $2$

Answer 1

Answer: (C)

$4 \alpha^{2}+2 \alpha-1=0$
$2(f(\alpha)+(\alpha))=2\left(4 \alpha^{3}-3 \alpha+1+\alpha\right)$
$=2\left(\alpha\left(4 \alpha^{2}\right)+(1-2 \alpha)\right)$
$=2(\alpha(1-2 \alpha)+(1-2 \alpha))$
$=2\left(\alpha-2 \alpha^{2}+1-2 \alpha\right)$
$=2\left(-2 \alpha^{2}-\alpha+1\right)$
$=-4 \alpha^{2}-2 \alpha+2$
$=-1+2=1$