Question

If $\alpha$ is a root of the equation $4x^2+2x-1=0$, then other root of the equation is

• A. $4{\alpha }^3-\alpha$
• B. $4{\alpha }^2-3\alpha$
• C. $4{\alpha }^3-3\alpha$
• D. $4{\alpha }^2-\alpha$

Answer 1

Answer: (C)

$4{\alpha }^3-3\alpha$

$4x^2+2x-1=0$

$x=\frac{-2\pm \sqrt{(2{)}^2-4(4\left)\right(-1)}}{2\left(4\right)}=\frac{-1\pm \sqrt{5}}{4}$

$\frac{\sqrt{5}-1}{4}=\cos \left({72}^o\right)-\frac{(\sqrt{5}+1)}{4}=-\cos 36$

$-\cos 36=\cos (36+180)=\cos (216{)}^o$

$=\cos \left({3.72}^o\right)$

$\cos 3x=4{\cos }^{3}x-3\cos x$

If one root $\cos x=\alpha$ then

other root will be $4{\alpha }^3-3\alpha$.