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

  • Thread starter Thread starter kaliprasad
  • Start date Start date
  • Tags Tags
    Root
AI Thread Summary
The discussion revolves around finding the other root of the quadratic equation $4x^2+2x-1=0$ given one root as $\alpha$. Participants confirm that if one root is $\alpha$, the other root can be expressed as $4\alpha^3-3\alpha$. Multiple users contribute solutions and validate the relationship, emphasizing the connection between the roots derived from the equation's coefficients. The conversation highlights the mathematical reasoning behind the derivation of the second root based on the properties of quadratic equations. Overall, the thread effectively demonstrates the relationship between the roots of the given polynomial.
kaliprasad
Gold Member
MHB
Messages
1,333
Reaction score
0
If one root of $4x^2+2x-1= 0 $ be $\alpha$ then show that other root is $4\alpha^3-3\alpha$
 
Mathematics news on Phys.org
kaliprasad said:
If one root of $4x^2+2x-1= 0 $ be $\alpha$ then show that other root is $4\alpha^3-3\alpha$
my solution:
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
 
Albert said:
my solution:
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

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 $\alpha$
 
kaliprasad said:
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 $\alpha$
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
 
Last edited by a moderator:
My solution:

Let:

$$\alpha=\frac{-1\pm\sqrt{5}}{4}$$

Then:

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

Thus, when $\alpha$ is one root, $4\alpha^3-3\alpha$ is the other.
 
MarkFL said:
My solution:

Let:

$$\alpha=\frac{-1\pm\sqrt{5}}{4}$$

Then:

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

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

a typo :
(*)should be: $4\alpha^3-3\alpha
=\dfrac{-1\mp\sqrt{5}}{4}$
 
Albert said:
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. :)
 
kaliprasad said:
If one root of $4x^2+2x-1= 0 $ be $\alpha$ then show that other root is $4\alpha^3-3\alpha$

My solution:

$$x=\frac{-1+\sqrt{5}}{4}$$ 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 $$x=-\left(\frac{1+\sqrt{5}}{4}\right)$$ 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: $\cos 3\alpha=4\cos^3 \alpha -3\cos\alpha$
 
kaliprasad said:
If one root of $4x^2+2x-1= 0 $ be $\alpha$ then show that other root is $4\alpha^3-3\alpha$

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.
 
  • #10
three good answers markfl, anemone and greg and here is my answer.

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)
 
Back
Top