Mock IMO contest problem:

Solve $8 x^4-3 x^3-464 x^2-850 x+384=0$.

Observe! The best problem solver is also the best observer, train your eyes to see things others won't necessarily see or on first glance. The eyes are the windows to everything.

We need to exercise our mind by solving intriguing challenges as they can certainly keep our brain sharp. Sharp mindedness means the quickness of a person to interpret, understand, analyze, or pick up an idea adequately. It is a quality that enables you to find a solution to a problem very quickly or answer a question very quickly or learn very quickly. A person with sharp brain is a keen and quick observer.

First, Descartes' Rule of Signs suggests that we might have $2$ positive real and $2$ negative real roots.

Next, if we let $x=\dfrac{3}{8}$, then the first two terms will cancel each other out since:

$\begin{align*}8 x^4-3 x^3&=8 \left(\dfrac{3}{8}\right)^4-3 \left(\dfrac{3}{8}\right)^3\\&=\dfrac{3^4}{8^3}-\dfrac{3^4}{8^3}\\&=0\end{align*}$

Also,

$\begin{align*}464 x^2+850 x-384&=2(232x^2-425x+192)\\&=2(29x+64)(8x-3)\end{align*}$

Putting them together, we see that we have:

$8 x^4-3 x^3-464 x^2-850 x+384=8 \left(\dfrac{3}{8}\right)^4-3 \left(\dfrac{3}{8}\right)^3-2(29x+64)(8x-3)=0$

That means $x=\dfrac{3}{8}$ is one of the positive root of the original equation.

Hence, performing the long division we get:

$8 x^4-3 x^3-464 x^2-850 x+384=0$

$(8x-3)(x^3-58 x-128)=0$

Ah! The second factor is easy to solve, since we could approach it using the Cardano's cubic formula.

Using the Cardano's cubic formula where it says if we have the cubic equation $x^3+ax^2+bx+c=0$, the solutions for $x$ are hence

$x_1=-\dfrac{1}{3}a+(S+T)$,

$x_2=-\dfrac{1}{3}a-\dfrac{1}{2}(S+T)+\dfrac{i\sqrt{3}}{2}(S-T)$ and

$x_3=-\dfrac{1}{3}a-\dfrac{1}{2}(S+T)-\dfrac{i\sqrt{3}}{2}(S-T)$

where

$Q=\dfrac{3b-a^2}{9},\,R=\dfrac{9ba-27c-2a^3}{54}$,

$D=Q^3+R^2$,

$S=(R+\sqrt{D})^{\frac{1}{3}}$, $T=(R-\sqrt{D})^{\frac{1}{3}}$.

In our case, we have:

$Q=\dfrac{3(-58)-0^2}{9}=-\dfrac{58}{3},\,R=\dfrac{9(-58)(0)-27(-128)-2(0)^3}{54}=64$,

$D=Q^3+R^2=\left(-\dfrac{58}{3}\right)^3+64^2=-\dfrac{84520}{27}$

Therefore

$S=\left(64+i\sqrt{\dfrac{84520}{27}}\right)^{\frac{1}{3}}$, and

$T=\left(64+i\sqrt{\dfrac{84520}{27}}\right)^{\frac{1}{3}}$

The solutions for $x$ are hence

$\small \begin{align*}x_1&=-\dfrac{1}{3}(0)+\left(64+i\sqrt{\dfrac{84520}{27}}\right)^{\frac{1}{3}}+\left(64-i\sqrt{\dfrac{84520}{27}}\right)^{\frac{1}{3}}\\&=\left(64+i\sqrt{\dfrac{84520}{27}}\right)^{\frac{1}{3}}+\left(64-i\sqrt{\dfrac{84520}{27}}\right)^{\frac{1}{3}}\end{align*}$

$\small \begin{align*}x_2&=-\dfrac{1}{3}(0)-\dfrac{1}{2}\left(\left(64+i\sqrt{\dfrac{84520}{27}}\right)^{\frac{1}{3}}+\left(64-i\sqrt{\dfrac{84520}{27}}\right)^{\frac{1}{3}}\right)+\dfrac{i\sqrt{3}}{2}\left(\left(64+i\sqrt{\dfrac{84520}{27}}\right)^{\frac{1}{3}}-\left(64-i\sqrt{\dfrac{84520}{27}}\right)^{\frac{1}{3}}\right)\\&=-\dfrac{1}{2}\left(\left(64+i\sqrt{\dfrac{84520}{27}}\right)^{\frac{1}{3}}+\left(64-i\sqrt{\dfrac{84520}{27}}\right)^{\frac{1}{3}}\right)+\dfrac{i\sqrt{3}}{2}\left(\left(64+i\sqrt{\dfrac{84520}{27}}\right)^{\frac{1}{3}}-\left(64-i\sqrt{\dfrac{84520}{27}}\right)^{\frac{1}{3}}\right)\end{align*}$

$\small \begin{align*}x_3&=-\dfrac{1}{2}\left(\left(64+i\sqrt{\dfrac{84520}{27}}\right)^{\frac{1}{3}}+\left(64-i\sqrt{\dfrac{84520}{27}}\right)^{\frac{1}{3}}\right)-\dfrac{i\sqrt{3}}{2}\left(\left(64+i\sqrt{\dfrac{84520}{27}}\right)^{\frac{1}{3}}-\left(64-i\sqrt{\dfrac{84520}{27}}\right)^{\frac{1}{3}}\right)\end{align*}$

To conclude, the solutions for solving the equation $8 x^4-3 x^3-464 x^2-850 x+384=0$ are the above three $x$ and $x=\dfrac{3}{8}$.

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