MHB Can You Match Constants to This Cubic Polynomial?

[anemone]

Find the constants $$a,\;b, \;c,\; d$$ such that

$$4x^3-3x+\frac{\sqrt{3}}{2}=a(x-b)(x-c)(x-d)$$.

 

[MarkFL]

Using the triple-angle identity for cosine:

(1) $\displaystyle \cos(3\theta)=4\cos^3(\theta)-3\cos(\theta)$

we may solve the cubic equation:

(2) $\displaystyle 8x^3-6x+\sqrt{3}=0$

To transform equation (2) into a form where the stated identity (1) is useful, we make the substitution $x=a\cos(\theta)$, where $a$ is a constant to be determined. With this substitution, equation (2) can be written:

(3) $\displaystyle 8a^3\cos^3(\theta)-6a\cos(\theta)=-\sqrt{3}$

In equation (3), the coefficient of $\cos^3(\theta)$ is $8a^3$. Since we want this coefficient to be $4$ [as it is in equation (1)], we divide both sides of equation (3) by $2a^3$ to obtain:

(4) $\displaystyle 4\cos^3(\theta)-\frac{3}{a^2}\cos(\theta)=-\frac{\sqrt{3}}{2a^3}$

Next, a comparison of equations (4) and (1) leads us to require that $\displaystyle \frac{3}{a^2}=3$. Thus, $a=\pm1$. For convenience, we choose $a=1$; equation (4) then becomes:

(5) $\displaystyle 4\cos^3(\theta)-3\cos(\theta)=-\frac{\sqrt{3}}{2}$

Comparing equation (5) with the identity in (1) leads us to the equation:

$\displaystyle \cos(3\theta)=-\frac{\sqrt{3}}{2}$

The solutions here are of the form:

$\displaystyle 3\theta=\frac{\pi}{6}(12k+5),\,\frac{\pi}{6}(12k+7)$

$\displaystyle \theta=\frac{\pi}{18}(12k+5),\,\frac{\pi}{18}(12k+7)$

Only 3 of these angles yield distinct values of $\cos(\theta)$, namely:

$\displaystyle \theta=\frac{5\pi}{18},\,\frac{7\pi}{18},\,\frac{17\pi}{18}$

Thus, the solutions of the equation in (2) are:

$\displaystyle x=\cos\left(\frac{5\pi}{18} \right),\,\cos\left(\frac{7\pi}{18} \right),\,\cos\left(\frac{17\pi}{18} \right)$

Hence, $a=4$, and $(b,c,d)$ can be any of the six permutations of the 3 roots listed above.

 

[anemone]

Hi MarkFL, thanks for participating and your answer is correct!:D

Hey, I'm impressed at how fast you were in replying to this problem!

 

[MarkFL]

I struggled at first with Vieta, but then I recalled that this is quite similar to the High School POTW that I submitted for use on 9 June, and so I copied and pasted the solution I had provided, made a few changes, and had it solved. (Angel)

 

[Opalg]

Find the constants $$a,\;b, \;c,\; d$$ such that

$$4x^3-3x+\frac{\sqrt{3}}{2}=a(x-b)(x-c)(x-d)$$.

Clearly $a=4$. For the rest ... [sp]$\cos(3\theta) = 4\cos^3\theta - 3\cos\theta$, so if we put $x=\cos\theta$ then the equation $4x^3-3x+\tfrac{\sqrt{3}}{2}=0$ becomes $\cos(3\theta) = -\tfrac{\sqrt3}2 = \cos150^\circ$, with solutions $\theta = 50^\circ,\;170^\circ,\;290^\circ$ or, if you prefer, $50^\circ,\;70^\circ,\;170^\circ$. Thus we can take $b= \cos50^\circ$, $c = \cos70^\circ$, $d = -\cos10^\circ$.[/sp]

Edit. Sorry! I took so long writing this that I failed to see that Mark had already replied.

 

[kaliprasad]

I had solve it at Fun with maths: Slightly hard cubic eqution

 

[Prove It]

Find the constants $$a,\;b, \;c,\; d$$ such that

$$4x^3-3x+\frac{\sqrt{3}}{2}=a(x-b)(x-c)(x-d)$$.

[math]\displaystyle \begin{align*} a\left( x-b \right) \left( x-c\right) \left( x-d \right) = a \, x^3 - \left( a\,b + a\,c + a\,d \right) x^2 + \left( a\,b\,c + a\,b\,d + a\,c\,d \right) x - a\,b\,c\,d \end{align*}[/math]

Equating the like powers of x we find

[math]\displaystyle \begin{align*} a &= 4 \\ \\ - \left( 4b + 4c + 4d \right) &= 0 \\ b + c + d &= 0 \\ \\ 4\,b\,c + 4\,b\,d + 4\,c\,d &= -3 \\ b\,c + b\,d + c\,d &= -\frac{3}{4} \\ \\ 4\,b\,c\,d &= \frac{\sqrt{3}}{2} \\ b\,c\,d &= \frac{\sqrt{3}}{8} \end{align*}[/math]

Rearranging the first equation we have [math]\displaystyle \begin{align*} b = -c - d \end{align*}[/math] and substituting into the second equation we find

[math]\displaystyle \begin{align*} \left( - c - d \right) c + \left( -c - d \right) d + c\,d &= -\frac{3}{4} \\ -c^2 - c\,d - c\,d - d^2 + c\,d &= -\frac{3}{4} \\ c^2 + c\,d + d^2 &= \frac{3}{4} \\ c^2 + c\,d + \left( \frac{d}{2} \right) ^2 - \left( \frac{d}{2} \right) ^2 + d^2 &= \frac{3}{4} \\ \left( c + \frac{d}{2} \right) ^2 + \frac{3d^2}{4} &= \frac{3}{4} \\ \left( c + \frac{d}{2} \right) ^2 &= \frac{ 3 - 3d }{4} \\ c + \frac{d}{2} &= \frac{ \pm \sqrt{ 3 - 3d }}{2} \\ c &= \frac{-d \pm \sqrt{ 3 - 3d } }{2} \end{align*}[/math]

and so [math]\displaystyle \begin{align*} b = - \left( \frac{-d \pm \sqrt{ 3 - 3d }}{2} \right) - d = \frac{ - d \mp \sqrt{ 3 - 3d }}{2} \end{align*}[/math]

Substituting into the final equation, we find

[math]\displaystyle \begin{align*} \left( \frac{ - d \mp \sqrt{ 3 - 3d }}{2} \right) \left( \frac{-d \pm \sqrt{ 3 - 3d } }{2} \right) d &= \frac{\sqrt{3}}{8} \\ \left( -d \mp \sqrt{ 3 - 3d} \right) \left( -d \pm \sqrt{ 3 - 3d } \right) d &= \frac{\sqrt{3}}{2} \\ \left[ d^2 - \left( 3 - 3d \right) \right] d &= \frac{\sqrt{3}}{2} \\ d^3 + 3d^2 - 3d - \frac{\sqrt{3}}{2} &= 0 \end{align*}[/math]

Now making the change of variable [math]\displaystyle \begin{align*} u = d + 1 \end{align*}[/math], we find

[math]\displaystyle \begin{align*} \left( u - 1 \right) ^3 + 3 \left( u - 1 \right) ^2 - 3 \left( u - 1 \right) - \frac{\sqrt{3}}{2} &= 0 \\ u^3 - 3u^2 + 3u - 1 + 3u^2 - 6u + 3 - 3u + 3 - \frac{\sqrt{3}}{2} &= 0 \\ u^3 - 6u - \frac{\sqrt{3} - 10}{2} &= 0 \end{align*}[/math]

and now applying the Cubic Formula, we have

[math]\displaystyle \begin{align*} u &= \sqrt[3]{ \frac{\frac{\sqrt{3} - 10}{2} + \sqrt{\left( \frac{\sqrt{3} - 10}{2} \right) ^2 - \frac{4 \left( - 6 \right) ^3}{27} }}{2} } + \sqrt[3]{ \frac{\frac{\sqrt{3} - 10}{2} - \sqrt{\left( \frac{\sqrt{3} - 10}{2} \right) ^2 - \frac{4 \left( - 6 \right) ^3}{27} }}{2} } \\ &= \sqrt[3]{ \frac{\sqrt{3} - 10 + \sqrt{ 231 - 20\sqrt{3} }}{4} } + \sqrt[3]{ \frac{\sqrt{3} - 10 - \sqrt{ 231 - 20\sqrt{3} }}{4} } \\ d + 1 &= \sqrt[3]{ \frac{\sqrt{3} - 10 + \sqrt{ 231 - 20\sqrt{3} }}{4} } + \sqrt[3]{ \frac{\sqrt{3} - 10 - \sqrt{ 231 - 20\sqrt{3} }}{4} } \\ d &= \sqrt[3]{ \frac{\sqrt{3} - 10 + \sqrt{ 231 - 20\sqrt{3} }}{4} } + \sqrt[3]{ \frac{\sqrt{3} - 10 - \sqrt{ 231 - 20\sqrt{3} }}{4} } - 1 \end{align*}[/math]

Back-substituting will enable us to evaluate b and c.

 

[I like Serena]

I have always liked Cardano's method that I learned a long long time ago from a book of my father. Since I still have fond sentimental memories of it, I'm presenting it as well.

We can write the cubic expression in equation form as:
$$x^3 - \frac 3 4 x + \frac {\sqrt 3} 8 = 0 \qquad (1)$$
We substitute
$$x=y+z \qquad\qquad (2)$$
giving us a free choice for either $y$ or $z$:
\begin{array}{lcl}
(y+z)^3 - \frac 3 4 (y+z) + \frac {\sqrt 3} 8
&=&(y^3+z^3) + (3y^2z+3yz^2) - \frac 3 4 (y+z) + \frac {\sqrt 3} 8 = 0 \\
&=&(y^3+z^3) + \left(3yz - \frac 3 4\right)(y+z) + \frac {\sqrt 3} 8
\end{array}
Now we make our choice for z such that $3yz - \frac 3 4 = 0$, which will make the second term vanish.
That is, we substitute:
$$z = \frac 1 {4y} \qquad\qquad (3)$$
The result is:
\begin{array}{lcl}
(y^3+\frac 1 {4^3 y^3}) + \frac {\sqrt 3} 8 &=& 0 \\
y^6 + \frac {\sqrt 3} 8 y^3 + \frac 1 {4^3} &=& 0
\end{array}
Solving the equivalent quadratic equation gives:
$$y^3 = \frac 1 8 \left(-\frac 12 \sqrt 3 \pm \frac 1 2 i\right) = \frac 1 8 \exp\left(\pm \frac 5 6 \pi i\right)$$
The corresponding solutions are:
$$y=\frac 1 2 \exp\left(\pm \frac 5 {18} \pi i\right), \quad \frac 1 2 \exp\left(\pm \frac {17} {18} \pi i\right), \quad\frac 1 2 \exp\left(\pm \frac {29} {18} \pi i\right)$$
Substituting in (3) gives us:
$$z=\frac 1 2 \exp\left(\mp \frac 5 {18} \pi i\right), \quad\frac 1 2 \exp\left(\mp \frac {17} {18} \pi i\right), \quad\frac 1 2 \exp\left(\mp \frac {29} {18} \pi i\right)$$
In other words, in each case we have $z=\bar y$.
As a result, from (2) we find $x=y+z=y+\bar y=2\Re (y)$, meaning:
$$x=\cos\left( \frac 5 {18} \pi\right), \quad\cos\left( \frac {17} {18} \pi\right), \quad\cos\left( \frac {29} {18} \pi\right)$$

Therefore the solution is
$$a=4, \quad b= \cos\left( \frac 5 {18} \pi\right), \quad c=\cos\left( \frac {17} {18} \pi\right), \quad d=\cos\left( \frac {29} {18} \pi\right). \qquad \blacksquare$$

 

[anemone]

Clearly $a=4$. For the rest ... [sp]$\cos(3\theta) = 4\cos^3\theta - 3\cos\theta$, so if we put $x=\cos\theta$ then the equation $4x^3-3x+\tfrac{\sqrt{3}}{2}=0$ becomes $\cos(3\theta) = -\tfrac{\sqrt3}2 = \cos150^\circ$, with solutions $\theta = 50^\circ,\;170^\circ,\;290^\circ$ or, if you prefer, $50^\circ,\;70^\circ,\;170^\circ$. Thus we can take $b= \cos50^\circ$, $c = \cos70^\circ$, $d = -\cos10^\circ$.[/sp]

Edit. Sorry! I took so long writing this that I failed to see that Mark had already replied.

Hi Opalg, don't be sorry because I bet you have no idea how much I hope you would reply to all of my challenge problems!:D

I had solve it at Fun with maths: Slightly hard cubic eqution

Hi kaliprasad, hmm...so, that blog is yours? Thanks for sharing it with us and I've checked it out and found that you have quite a collection of interesting problems in your blog! My hearty wishes for your blogging success!:)

[math]\displaystyle \begin{align*} a\left( x-b \right) \left( x-c\right) \left( x-d \right) = a \, x^3 - \left( a\,b + a\,c + a\,d \right) x^2 + \left( a\,b\,c + a\,b\,d + a\,c\,d \right) x - a\,b\,c\,d \end{align*}[/math]

Equating the like powers of x we find

[math]\displaystyle \begin{align*} a &= 4 \\ \\ - \left( 4b + 4c + 4d \right) &= 0 \\ b + c + d &= 0 \\ \\ 4\,b\,c + 4\,b\,d + 4\,c\,d &= -3 \\ b\,c + b\,d + c\,d &= -\frac{3}{4} \\ \\ 4\,b\,c\,d &= \frac{\sqrt{3}}{2} \\ b\,c\,d &= \frac{\sqrt{3}}{8} \end{align*}[/math]

Rearranging the first equation we have [math]\displaystyle \begin{align*} b = -c - d \end{align*}[/math] and substituting into the second equation we find

[math]\displaystyle \begin{align*} \left( - c - d \right) c + \left( -c - d \right) d + c\,d &= -\frac{3}{4} \\ -c^2 - c\,d - c\,d - d^2 + c\,d &= -\frac{3}{4} \\ c^2 + c\,d + d^2 &= \frac{3}{4} \\ c^2 + c\,d + \left( \frac{d}{2} \right) ^2 - \left( \frac{d}{2} \right) ^2 + d^2 &= \frac{3}{4} \\ \left( c + \frac{d}{2} \right) ^2 + \frac{3d^2}{4} &= \frac{3}{4} \\ \left( c + \frac{d}{2} \right) ^2 &= \frac{ 3 - 3d }{4} \\ c + \frac{d}{2} &= \frac{ \pm \sqrt{ 3 - 3d }}{2} \\ c &= \frac{-d \pm \sqrt{ 3 - 3d } }{2} \end{align*}[/math]

and so [math]\displaystyle \begin{align*} b = - \left( \frac{-d \pm \sqrt{ 3 - 3d }}{2} \right) - d = \frac{ - d \mp \sqrt{ 3 - 3d }}{2} \end{align*}[/math]

Substituting into the final equation, we find

[math]\displaystyle \begin{align*} \left( \frac{ - d \mp \sqrt{ 3 - 3d }}{2} \right) \left( \frac{-d \pm \sqrt{ 3 - 3d } }{2} \right) d &= \frac{\sqrt{3}}{8} \\ \left( -d \mp \sqrt{ 3 - 3d} \right) \left( -d \pm \sqrt{ 3 - 3d } \right) d &= \frac{\sqrt{3}}{2} \\ \left[ d^2 - \left( 3 - 3d \right) \right] d &= \frac{\sqrt{3}}{2} \\ d^3 + 3d^2 - 3d - \frac{\sqrt{3}}{2} &= 0 \end{align*}[/math]

Now making the change of variable [math]\displaystyle \begin{align*} u = d + 1 \end{align*}[/math], we find

[math]\displaystyle \begin{align*} \left( u - 1 \right) ^3 + 3 \left( u - 1 \right) ^2 - 3 \left( u - 1 \right) - \frac{\sqrt{3}}{2} &= 0 \\ u^3 - 3u^2 + 3u - 1 + 3u^2 - 6u + 3 - 3u + 3 - \frac{\sqrt{3}}{2} &= 0 \\ u^3 - 6u - \frac{\sqrt{3} - 10}{2} &= 0 \end{align*}[/math]

and now applying the Cubic Formula, we have

[math]\displaystyle \begin{align*} u &= \sqrt[3]{ \frac{\frac{\sqrt{3} - 10}{2} + \sqrt{\left( \frac{\sqrt{3} - 10}{2} \right) ^2 - \frac{4 \left( - 6 \right) ^3}{27} }}{2} } + \sqrt[3]{ \frac{\frac{\sqrt{3} - 10}{2} - \sqrt{\left( \frac{\sqrt{3} - 10}{2} \right) ^2 - \frac{4 \left( - 6 \right) ^3}{27} }}{2} } \\ &= \sqrt[3]{ \frac{\sqrt{3} - 10 + \sqrt{ 231 - 20\sqrt{3} }}{4} } + \sqrt[3]{ \frac{\sqrt{3} - 10 - \sqrt{ 231 - 20\sqrt{3} }}{4} } \\ d + 1 &= \sqrt[3]{ \frac{\sqrt{3} - 10 + \sqrt{ 231 - 20\sqrt{3} }}{4} } + \sqrt[3]{ \frac{\sqrt{3} - 10 - \sqrt{ 231 - 20\sqrt{3} }}{4} } \\ d &= \sqrt[3]{ \frac{\sqrt{3} - 10 + \sqrt{ 231 - 20\sqrt{3} }}{4} } + \sqrt[3]{ \frac{\sqrt{3} - 10 - \sqrt{ 231 - 20\sqrt{3} }}{4} } - 1 \end{align*}[/math]

Back-substituting will enable us to evaluate b and c.

Hi Prove It, thanks for providing us this neat and well written solution and I appreciate that you took the time to participate in this problem.:)

I have always liked Cardano's method that I learned a long long time ago from a book of my father. Since I still have fond sentimental memories of it, I'm presenting it as well.

We can write the cubic expression in equation form as:
$$x^3 - \frac 3 4 x + \frac {\sqrt 3} 8 = 0 \qquad (1)$$
We substitute
$$x=y+z \qquad\qquad (2)$$
giving us a free choice for either $y$ or $z$:
\begin{array}{lcl}
(y+z)^3 - \frac 3 4 (y+z) + \frac {\sqrt 3} 8
&=&(y^3+z^3) + (3y^2z+3yz^2) - \frac 3 4 (y+z) + \frac {\sqrt 3} 8 = 0 \\
&=&(y^3+z^3) + \left(3yz - \frac 3 4\right)(y+z) + \frac {\sqrt 3} 8
\end{array}
Now we make our choice for z such that $3yz - \frac 3 4 = 0$, which will make the second term vanish.
That is, we substitute:
$$z = \frac 1 {4y} \qquad\qquad (3)$$
The result is:
\begin{array}{lcl}
(y^3+\frac 1 {4^3 y^3}) + \frac {\sqrt 3} 8 &=& 0 \\
y^6 + \frac {\sqrt 3} 8 y^3 + \frac 1 {4^3} &=& 0
\end{array}
Solving the equivalent quadratic equation gives:
$$y^3 = \frac 1 8 \left(-\frac 12 \sqrt 3 \pm \frac 1 2 i\right) = \frac 1 8 \exp\left(\pm \frac 5 6 \pi i\right)$$
The corresponding solutions are:
$$y=\frac 1 2 \exp\left(\pm \frac 5 {18} \pi i\right), \quad \frac 1 2 \exp\left(\pm \frac {17} {18} \pi i\right), \quad\frac 1 2 \exp\left(\pm \frac {29} {18} \pi i\right)$$
Substituting in (3) gives us:
$$z=\frac 1 2 \exp\left(\mp \frac 5 {18} \pi i\right), \quad\frac 1 2 \exp\left(\mp \frac {17} {18} \pi i\right), \quad\frac 1 2 \exp\left(\mp \frac {29} {18} \pi i\right)$$
In other words, in each case we have $z=\bar y$.
As a result, from (2) we find $x=y+z=y+\bar y=2\Re (y)$, meaning:
$$x=\cos\left( \frac 5 {18} \pi\right), \quad\cos\left( \frac {17} {18} \pi\right), \quad\cos\left( \frac {29} {18} \pi\right)$$

Therefore the solution is
$$a=4, \quad b= \cos\left( \frac 5 {18} \pi\right), \quad c=\cos\left( \frac {17} {18} \pi\right), \quad d=\cos\left( \frac {29} {18} \pi\right). \qquad \blacksquare$$

Hi I like Serena, WOW! Thank you kindly for this piece of well thought out strategic plan to tackle this cubic function!

I am so happy that all of you so generously shared the different approaches to solve this problem with me!

I love you guys!(Nerd)

 

[kaliprasad]

Hi kaliprasad, hmm...so, that blog is yours? Thanks for sharing it with us and I've checked it out and found that you have quite a collection of interesting problems in your blog! My hearty wishes for your blogging success!:)

Yes. The blog is mine and thanks for wishing me blogging success!