Find other root of equation $4x^2+ 2x –1 = 0$.

Question

If $k$ be a root of the equation $4x^2+ 2x -1 = 0$ then prove that $4k^3-3k$ is the other root.

I simply get the other root as $-\dfrac{1}{4k}$.

As product of roots is $-\dfrac{1}{4}$.

But how to prove that?

Answer 1

Let the other root be given by $\alpha $. Then we have, $$ k+\alpha =- 0.5 \text { and } k\alpha =-0.25$$ Then, $k-\alpha = \sqrt { (k+\alpha)^2-4k\alpha} = \sqrt {0.25+1} =\frac {\sqrt {5}}{2} $

Thus, $k = \frac {\sqrt {5}-1}{4} $ and $\alpha = -\frac {\sqrt {5}+1}{4} $. We know that $\sin 18^\circ = \frac {\sqrt {5}-1}{4} = k $ and $\sin 54^\circ = \frac {\sqrt {5}+1}{4} = -\alpha = \sin (3\times 18^\circ) = -4\sin^3 18^\circ + 3\sin 18^\circ = -4k^3+3k $ giving us $\alpha =4k^3-3k $ proving the result. Hope it helps.

Answer 2

By Vieta $\,k'\! = $ other root $\iff kk' = -\frac{1}4\iff 4kk'+1 = 0,\,$ which is true for $\,k' = 4k^3-3k$

$$ \begin{align} 4k\,(\overbrace{4k^3-3k}^{\large k'})+1 &=\, (4k^2-1)^2-\,(2k)^2\\[.2em] &=\, (\color{#c00}{4k^2-1+2k})\ (4k^2-1-2k)\\[.2em] &=\,\ \color{#c00}0 \end{align}$$

Answer 3

Consider $2x=T$.
You can then solve it quickly.