Showing the Other Root of $4x^2+2x-1= 0$ Given $\alpha$

[kaliprasad]

If one root of $4x^2+2x-1= 0 $ be $\alpha$ then show that other root is $4\alpha^3-3\alpha$

 

[Albert1]

If one root of $4x^2+2x-1= 0 $ be $\alpha$ then show that other root is $4\alpha^3-3\alpha$

my solution:

Spoiler

for $\alpha $ is a root ,we must have $4\alpha^2+2\alpha -1=0----(1)$
let $\alpha +4\alpha^3-3\alpha=x----(2)$
and $\alpha \times (4\alpha^3-3\alpha)=4\alpha^4-3\alpha^2=y----(3)$
$(2)\times \alpha \rightarrow 4\alpha^4-2\alpha^2=x\alpha----(4)$
$(4)-(3):\alpha^2=x\alpha -y \rightarrow 4\alpha^2-4x\alpha+4y=0----(5)$
compare$(1)\,\, and\,\, (5)$,we have $x=\dfrac{-1}{2},y=\dfrac {-1}{4}$
and the proof is finished

 

[kaliprasad]

my solution:

Spoiler

for $\alpha $ is a root ,we must have $4\alpha^2+2\alpha -1=0----(1)$
let $\alpha +4\alpha^3-3\alpha=x----(2)$
and $\alpha \times (4\alpha^3-3\alpha)=4\alpha^4-3\alpha^2=y----(3)$
$(2)\times \alpha \rightarrow 4\alpha^4-2\alpha^2=x\alpha----(4)$
$(4)-(3):\alpha^2=x\alpha -y \rightarrow 4\alpha^2-4x\alpha+4y=0----(5)$
compare$(1)\,\, and\,\, (5)$,we have $x=\dfrac{-1}{2},y=\dfrac {-1}{4}$
and the proof is finished

Spoiler

this is not proof as you are asuming the result and based on it you prove it you have assumed that (1) is the only equation satisfied by $\alpha$

 

[Albert1]

Spoiler

this is not proof as you are asuming the result and based on it you prove it you have assumed that (1) is the only equation satisfied by $\alpha$

Spoiler

I prove $\alpha \,\, and \,\,4\alpha^3-3\alpha$ satisfy Vieta's formulas
if $\alpha $ is a root ,and $4\alpha^3-3\alpha$ is another root ,then their sum and product must satisfy Vieta's formulas
and more $4x^2+2x-1=0 $ has exactly only two roots

 

[MarkFL]

My solution:

Spoiler

Let:

\(\displaystyle \alpha=\frac{-1\pm\sqrt{5}}{4}\)

Then:

\(\displaystyle 4\alpha^3-3\alpha=4\left(\frac{-1\pm\sqrt{5}}{4}\right)^3-3\left(\frac{-1\pm\sqrt{5}}{4}\right)=\frac{-16\pm8\sqrt{5}+12\mp12\sqrt{5}}{16}=\frac{-1\mp\sqrt{5}}{4}\)

Thus, when $\alpha$ is one root, $4\alpha^3-3\alpha$ is the other.

 

[Albert1]

My solution:

Spoiler

Let:

\(\displaystyle \alpha=\frac{-1\pm\sqrt{5}}{4}\)

Then:

\(\displaystyle 4\alpha^3-3\alpha=4\left(\frac{-1\pm\sqrt{5}}{4}\right)^3-3\left(\frac{-1\pm\sqrt{5}}{4}\right)=\frac{-16\pm8\sqrt{5}+12\mp12\sqrt{5}}{16}=\frac{-1\mp\sqrt{5}}{5}--(*)\)

Thus, when $\alpha$ is one root, $4\alpha^3-3\alpha$ is the other.

Spoiler

a typo :
(*)should be: $4\alpha^3-3\alpha
=\dfrac{-1\mp\sqrt{5}}{4}$

 

[MarkFL]

Spoiler

a typo :
(*)should be: $4\alpha^3-3\alpha
=\dfrac{-1\mp\sqrt{5}}{4}$

Fixed...and I hid the information in your post regarding a solution using the spoiler tags as a courtesy to the community. :)

 

[anemone]

If one root of $4x^2+2x-1= 0 $ be $\alpha$ then show that other root is $4\alpha^3-3\alpha$

My solution:

Spoiler

\(\displaystyle x=\frac{-1+\sqrt{5}}{4}\) is a solution for $4x^2+2x-1= 0$ and it's also known to be $\cos 72^\circ$(sum 36+36). The other solution for $4x^2+2x-1= 0$ is \(\displaystyle x=-\left(\frac{1+\sqrt{5}}{4}\right)\) and it's known to be $-\cos 36^\circ=\cos (180^\circ+36^\circ)=\cos 3(72^\circ)$.

We therefore can conclude that if one of the roots for $4x^2+2x-1= 0$ is $\alpha$, which is $\cos 72^\circ$, the other root, $\cos 3(72^\circ)$ will be $4\alpha^3-3\alpha$, which abides by the triple angle formula for cosine function: $\cos 3\alpha=4\cos^3 \alpha -3\cos\alpha$

 

[Greg]

If one root of $4x^2+2x-1= 0 $ be $\alpha$ then show that other root is $4\alpha^3-3\alpha$

Spoiler

I know that $\cos(36^\circ)=\dfrac{\sqrt5+1}{4}$. By the quadratic relation, the two roots of the given quadratic are

$$\dfrac{\sqrt5-1}{4}$$ and $$-\dfrac{\sqrt5+1}{4}$$

so we have

$$\cos(36^\circ)-\dfrac{\sqrt5+1}{2}=-\dfrac{\sqrt5+1}{4}$$

$$\cos(36^\circ)-2\cos(36^\circ)=-\dfrac{\sqrt5+1}{4}$$

$$-\cos(36^\circ)=\cos(216^\circ)=-\dfrac{\sqrt5+1}{4}$$

thus the other root is $\cos(216^\circ)$.

Now,

$$4\cos^3(x)-3\cos(x)=\cos(3x)$$

so, if $\cos(72)=\dfrac{\sqrt5-1}{4}$, we are done.

$$\cos(72^\circ)=2\cos^2(36^\circ)-1=2\left(\dfrac{\sqrt5+1}{4}\right)^2-1=\dfrac{\sqrt5-1}{4}$$

as required.

Thanks to anemone for inspiration.

 

[kaliprasad]

three good answers markfl, anemone and greg and here is my answer.

Spoiler

We have $4\alpha^2+2\alpha-1= 0\cdots(1) $
We need to prove
$4(4\alpha^3-3\alpha)^2+2(4\alpha^3-3\alpha)-1= 0 $
we have $4\alpha^2+2\alpha-1= 0$
Hence $4\alpha^3+2\alpha^2-\alpha= 0$
$4\alpha^3 = -2\alpha^2 + \alpha$
or $4\alpha^3 -3\alpha = - 2\alpha^2 - 2 \alpha= - \frac{1}{2}( 4\alpha^2 + 4\alpha)$
$= - \frac{1}{2}( 1 - 2\alpha + 4\alpha)= - \frac{1}{2}( 1 + 2\alpha) \dots(2)$
Hence $4(4\alpha^3-3\alpha)^2+2(4\alpha^3-3\alpha)-1= 0 $
= $4(\frac{1}{2}( 1 + 2\alpha))^2- 2 (\frac{1}{2}( 1 + 2\alpha)-1$ using (2)
= $( 1 + 2\alpha))^2- ( 1 + 2\alpha)-1$
= $1 + 4 \alpha + 4\alpha ^2 - 1 -2 \alpha -1$
= $4\alpha ^2 + 2 \alpha -1= 0$ from (1)