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)

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