Question

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

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

Answer 1

Answer: (B)

Given equation is $4x^2+2x-1=0$
Let $\alpha ,\beta$ be roots of this equation
Then $\alpha +\beta =\frac{-1}{2}$ and $\alpha \beta =\frac{-1}{4}$
Also ${\left(\alpha -\beta \right)}^2={\left(\alpha +\beta \right)}^2-4\alpha \beta$
$=\frac{1}{4}+1=\frac{5}{4}$
$\Rightarrow \alpha -\beta =\frac{\sqrt{5}}{2}$
$\Rightarrow \beta =\frac{-1-\sqrt{5}}{2\times 2}=\frac{-1-\sqrt{5}}{4}$
Since $\alpha \beta =\frac{-1}{4}$
$\therefore \alpha =\frac{1}{1+\sqrt{5}}\times \frac{\sqrt{5}-1}{\sqrt{5}-1}=\frac{\sqrt{5}-1}{4}$
Now, $4{\alpha }^3-3\alpha =\frac{{\left(\sqrt{5}-1\right)}^3}{16}-\frac{3\left(\sqrt{5}-1\right)}{4}$
$=\frac{{\left(\sqrt{5}\right)}^3-1-3\sqrt{5}\left(\sqrt{5}-1\right)-12\sqrt{5}+12}{16}$
$=\frac{5\sqrt{5}-1-15+3\sqrt{5}-12\sqrt{5}+12}{16}$
$=\frac{-\sqrt{5}-1}{4}=\beta$.