Question

lf $\alpha ,\beta (\alpha >\beta )$ are the roots of the equation $4x^2+2x-1=0$, then $\beta$ is equal to

• A. $-2\alpha -1$
• B. $4{\alpha }^2+\alpha -1$
• C. $4{\alpha }^3-3\alpha$
• D. $4{\alpha }^2-3\alpha$

Answer 1

Answer: (C)

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

Given equation is $4x^2+2x-1=0$

$\Rightarrow$ Either $x=\frac{-2-\sqrt{4+4\times 4}}{2\times 4}=\frac{\sqrt{5}-1}{4}=\sin {18}^{\circ }.\Rightarrow \alpha =\sin {18}^{\circ }.$

or $x=\frac{-2+\sqrt{4+4x}}{2}=-\frac{\sqrt{5}+1}{4}=-\sin {54}^{\circ }.\Rightarrow \beta =-\sin {54}^{\circ }.\Rightarrow \beta =-\sin 3\times {18}^{\circ }.=-(3\sin {18}^{\circ }-4{\sin }^3{18}^{\circ })$

Substitute for $\sin {18}^{\circ }=\alpha$ , we have

$\beta =-(3\alpha -4{\alpha }^3)=4{\alpha }^3-3\alpha$