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Q.
If $\alpha$ is one root of the equation $4x^{2} + 2x - 1 = 0$, then the other root is
Complex Numbers and Quadratic Equations
Solution:
Given equation is $4x^{2} + 2 x - 1 = 0$
Let $\alpha, \beta$ be roots of this equation
Then $\alpha+\beta=\frac{-1}{2}$ and $\alpha\beta=\frac{-1}{4}$
Also $\left(\alpha-\beta\right)^{2} =\left(\alpha+\beta\right)^{2}-4\alpha\beta$
$=\frac{1}{4}+1=\frac{5}{4}$
$\Rightarrow \, \alpha-\beta=\frac{\sqrt{5}}{2}$
$\Rightarrow \beta=\frac{-1-\sqrt{5}}{2\times2}=\frac{-1-\sqrt{5}}{4}$
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, $4\alpha^{3}-3\alpha=\frac{\left(\sqrt{5}-1\right)^{3}}{16}-\frac{3\left(\sqrt{5}-1\right)}{4}$
$=\frac{\left(\sqrt{5}\right)^{3} -1-3\sqrt{5}\left(\sqrt{5}-1\right)-12\sqrt{5}+12}{16}$
$=\frac{5\sqrt{5}-1-15+3\sqrt{5}-12\sqrt{5}+12}{16}$
$=\frac{-\sqrt{5}-1}{4}=\beta$.