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- Feb 14, 2012
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Find the constants \(\displaystyle a,\;b, \;c,\; d\) such that
\(\displaystyle 4x^3-3x+\frac{\sqrt{3}}{2}=a(x-b)(x-c)(x-d)\).
\(\displaystyle 4x^3-3x+\frac{\sqrt{3}}{2}=a(x-b)(x-c)(x-d)\).
Clearly $a=4$. For the rest ...Find the constants \(\displaystyle a,\;b, \;c,\; d\) such that
\(\displaystyle 4x^3-3x+\frac{\sqrt{3}}{2}=a(x-b)(x-c)(x-d)\).
\(\displaystyle \displaystyle \begin{align*} a\left( x-b \right) \left( x-c\right) \left( x-d \right) = a \, x^3 - \left( a\,b + a\,c + a\,d \right) x^2 + \left( a\,b\,c + a\,b\,d + a\,c\,d \right) x - a\,b\,c\,d \end{align*}\)Find the constants \(\displaystyle a,\;b, \;c,\; d\) such that
\(\displaystyle 4x^3-3x+\frac{\sqrt{3}}{2}=a(x-b)(x-c)(x-d)\).
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!Clearly $a=4$. For the rest ...$\cos(3\theta) = 4\cos^3\theta - 3\cos\theta$, 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 $b= \cos50^\circ$, $c = \cos70^\circ$, $d = -\cos10^\circ$.
Edit. Sorry! I took so long writing this that I failed to see that Mark had already replied.
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!I had solve it at Fun with maths: Slightly hard cubic eqution
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.\(\displaystyle \displaystyle \begin{align*} a\left( x-b \right) \left( x-c\right) \left( x-d \right) = a \, x^3 - \left( a\,b + a\,c + a\,d \right) x^2 + \left( a\,b\,c + a\,b\,d + a\,c\,d \right) x - a\,b\,c\,d \end{align*}\)
Equating the like powers of x we find
\(\displaystyle \displaystyle \begin{align*} a &= 4 \\ \\ - \left( 4b + 4c + 4d \right) &= 0 \\ b + c + d &= 0 \\ \\ 4\,b\,c + 4\,b\,d + 4\,c\,d &= -3 \\ b\,c + b\,d + c\,d &= -\frac{3}{4} \\ \\ 4\,b\,c\,d &= \frac{\sqrt{3}}{2} \\ b\,c\,d &= \frac{\sqrt{3}}{8} \end{align*}\)
Rearranging the first equation we have \(\displaystyle \displaystyle \begin{align*} b = -c - d \end{align*}\) and substituting into the second equation we find
\(\displaystyle \displaystyle \begin{align*} \left( - c - d \right) c + \left( -c - d \right) d + c\,d &= -\frac{3}{4} \\ -c^2 - c\,d - c\,d - d^2 + c\,d &= -\frac{3}{4} \\ c^2 + c\,d + d^2 &= \frac{3}{4} \\ c^2 + c\,d + \left( \frac{d}{2} \right) ^2 - \left( \frac{d}{2} \right) ^2 + d^2 &= \frac{3}{4} \\ \left( c + \frac{d}{2} \right) ^2 + \frac{3d^2}{4} &= \frac{3}{4} \\ \left( c + \frac{d}{2} \right) ^2 &= \frac{ 3 - 3d }{4} \\ c + \frac{d}{2} &= \frac{ \pm \sqrt{ 3 - 3d }}{2} \\ c &= \frac{-d \pm \sqrt{ 3 - 3d } }{2} \end{align*}\)
and so \(\displaystyle \displaystyle \begin{align*} b = - \left( \frac{-d \pm \sqrt{ 3 - 3d }}{2} \right) - d = \frac{ - d \mp \sqrt{ 3 - 3d }}{2} \end{align*}\)
Substituting into the final equation, we find
\(\displaystyle \displaystyle \begin{align*} \left( \frac{ - d \mp \sqrt{ 3 - 3d }}{2} \right) \left( \frac{-d \pm \sqrt{ 3 - 3d } }{2} \right) d &= \frac{\sqrt{3}}{8} \\ \left( -d \mp \sqrt{ 3 - 3d} \right) \left( -d \pm \sqrt{ 3 - 3d } \right) d &= \frac{\sqrt{3}}{2} \\ \left[ d^2 - \left( 3 - 3d \right) \right] d &= \frac{\sqrt{3}}{2} \\ d^3 + 3d^2 - 3d - \frac{\sqrt{3}}{2} &= 0 \end{align*}\)
Now making the change of variable \(\displaystyle \displaystyle \begin{align*} u = d + 1 \end{align*}\), we find
\(\displaystyle \displaystyle \begin{align*} \left( u - 1 \right) ^3 + 3 \left( u - 1 \right) ^2 - 3 \left( u - 1 \right) - \frac{\sqrt{3}}{2} &= 0 \\ u^3 - 3u^2 + 3u - 1 + 3u^2 - 6u + 3 - 3u + 3 - \frac{\sqrt{3}}{2} &= 0 \\ u^3 - 6u - \frac{\sqrt{3} - 10}{2} &= 0 \end{align*}\)
and now applying the Cubic Formula, we have
\(\displaystyle \displaystyle \begin{align*} u &= \sqrt[3]{ \frac{\frac{\sqrt{3} - 10}{2} + \sqrt{\left( \frac{\sqrt{3} - 10}{2} \right) ^2 - \frac{4 \left( - 6 \right) ^3}{27} }}{2} } + \sqrt[3]{ \frac{\frac{\sqrt{3} - 10}{2} - \sqrt{\left( \frac{\sqrt{3} - 10}{2} \right) ^2 - \frac{4 \left( - 6 \right) ^3}{27} }}{2} } \\ &= \sqrt[3]{ \frac{\sqrt{3} - 10 + \sqrt{ 231 - 20\sqrt{3} }}{4} } + \sqrt[3]{ \frac{\sqrt{3} - 10 - \sqrt{ 231 - 20\sqrt{3} }}{4} } \\ d + 1 &= \sqrt[3]{ \frac{\sqrt{3} - 10 + \sqrt{ 231 - 20\sqrt{3} }}{4} } + \sqrt[3]{ \frac{\sqrt{3} - 10 - \sqrt{ 231 - 20\sqrt{3} }}{4} } \\ d &= \sqrt[3]{ \frac{\sqrt{3} - 10 + \sqrt{ 231 - 20\sqrt{3} }}{4} } + \sqrt[3]{ \frac{\sqrt{3} - 10 - \sqrt{ 231 - 20\sqrt{3} }}{4} } - 1 \end{align*}\)
Back-substituting will enable us to evaluate b and c.
Hi I like Serena, WOW! Thank you kindly for this piece of well thought out strategic plan to tackle this cubic function!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
$$a=4, \quad b= \cos\left( \frac 5 {18} \pi\right), \quad c=\cos\left( \frac {17} {18} \pi\right), \quad d=\cos\left( \frac {29} {18} \pi\right). \qquad \blacksquare$$