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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?

egreg
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a25bedc5-3d09-41b8-82fb-ea6c353d75ae
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3 Answers3

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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.

  • For derivation of $\sin 18^\circ $ and related angles, see here. –  Jan 31 '17 at 14:27
  • A quibble: the problem stipulates that $k$ is a root of the quadratic. But this answer deduces, midway through, that it is a specific root. – Barry Cipra Jan 31 '17 at 16:24
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    Less of a quibble: The deduced value of $\alpha$ needs a minus sign. (As given, the two roots are not conjugates.). This casts doubt on the rest of the answer. – Barry Cipra Jan 31 '17 at 16:39
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By Vieta $\,k'\! = $ other root $\iff kk' = -\frac{1}4\iff 4kk'+1 = 0,\,$ which is true for $\,k' = 4k^3-3k$

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

Bill Dubuque
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    Note We could convert to a monic polynomial by changing variables $,X = 2x,\ K = 2k,\ K' = 2k',$ as in the AC-method. Then the above is $$\begin{align} \overbrace{K}^{\large 2k},(\overbrace{K^3-3K}^{\large 2k'})+1 &=, (K^2-1)^2-,K^2\[.2em] &=, (\color{#c00}{K^2-1+K})\ (K^2-1-K)\[.2em] &=,\ \color{#c00}0 \end{align}$$ – Bill Dubuque Jan 31 '17 at 15:53
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Consider $2x=T$.
You can then solve it quickly.

George Law
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KAYHAAN
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    How , we can solve it quickly – a25bedc5-3d09-41b8-82fb-ea6c353d75ae Jan 31 '17 at 18:45
  • using delta....... Easy peasy in case of a quadratic equation with a positive discriminate, the roots are real while a 0 discriminate indicates a single real root. A negative discriminant indicates imaginary (complex number format) roots.the discriminant of the equation a(x^2) + bx + c is (b^2) - 4ac. This is a straightforward way of calculating the discriminant. This same discriminant can be used to even find the roots of the quadratic equation. – KAYHAAN Feb 04 '17 at 16:37