# Polynomial equation

• Apr 9th 2013, 02:33 PM
Ippo
Polynomial equation
(x²+2)/2=2√(x³+1)

Have tried a lot of things, but I am pretty certain that I am making a mistake in my initial steps so I dont think
a copy of my tries would be of help.

• Apr 9th 2013, 03:00 PM
emakarov
Re: Polynomial equation
Have you tried denoting x^2 + 1 by t?
• Apr 9th 2013, 03:54 PM
Ippo
Re: Polynomial equation
not specifically x^2 + 1 , but i have tried the method , and after reducing the equation to x^4 - 16x^3 - 12 = 0 - I am blank .
Any way to get a Real number solution ?
• Apr 9th 2013, 04:14 PM
HallsofIvy
Re: Polynomial equation
First, that's not a polynomial equation. Second there is a missing "(" on the left. Is it supposed to be $(x^2+ 2)/2$?
And it is hard to read- that is $x^3+ 1$ on the right, not $x^2+ 1$ as emkarov seemed to think.

The first thing you should do is make it a polynomial equation by squaring both sides.
$\frac{x^4+ 4x^2+ 4}{4}= 4(x^3+ 1)$
$x^4+ 4x^2+ 4= 16x^3+ 16$
$x^4- 16x^3+ 4x^2- 12= 0$.

By the "rational root theorem" the only possible roots of that are the factor of 12: 1, -1, 2, -2, 4, -4, 8, -8, 12, and -12.

And unless I have done the arithemtic wrong none of those satisfy the equation so the equation has only irrational or complex roots.

Quote:

Originally Posted by Ippo
x²+2)/2=2√(x³+1)

Have tried a lot of things, but I am pretty certain that I am making a mistake in my initial steps so I dont think
a copy of my tries would be of help.

• Apr 9th 2013, 04:20 PM
Ippo
Re: Polynomial equation
what happened to the 4 in the denominator
• Apr 9th 2013, 04:22 PM
HallsofIvy
Re: Polynomial equation
I knew some one would catch me! It took me too long to correct it.
• Sep 5th 2013, 08:45 PM
thevinh
Re: Polynomial equation
$\text{Currently I cannot find a substitution method to solve this problem. However, it can be solved by brute force!}$

$x^2+2=4\sqrt{x^3+1}~\Leftrightarrow~x^4-16x^3+4x^2-12=0\qquad (1)$

$\text{There is no real rational solution to (1), therefore we have to solve the \emph{quartic equation} using \textbf{Ferrari's Method}}.$

$\text{Let}~ x=y-\frac{b}{4a}~\Rightarrow~x=y+4.~\text{This substitution will transform}~(1)~\text{into:}$

$y^4-92y^2-480y-716=0~\Leftrightarrow~y^4=92y^2+480y+716$

$\text{Add}~2zy^2+z^2~\text{to both sides of the equation, for some}~z\in R.~\text{The added terms will make the left side of the equation a perfect square.}$

$\left(y^2+z\right)^2=(2z+92)y^2+480y+716+z^2\qquad (2)$

$\text{The right side is a perfect square if its discriminant is equal to 0, since it is a quadratic equation in terms of}~y.$

$D=230400-4(2z+92)(z^2+716)=0~\Leftrightarrow~28800-(z+46)(z^2+716)=0$

$\Leftrightarrow~z^3+46z^3+716z+4136=0\qquad(3)$

$\text{Let}~z=u-\frac{b}{3a}~\Leftrightarrow~z=u-\frac{46}{3}\qquad (4)$

$(3)~\Leftrightarrow~u^3+\frac{32}{3}u+\frac{9920}{ 27}~\Leftrightarrow~27u^3+288u+9920=0$

$\Leftrightarrow~(3u+20)(9u^2-60u+496)=0~\Rightarrow~u=-\frac{20}{3}~\Rightarrow~z=-22$

$(2)~\Leftrightarrow~\left(y^2-22\right)^2=48y^2+480y+1200=48(y+5)^2$

$\Leftrightarrow~y^2-22=\pm4\sqrt{3}(y+5)~\Rightarrow~\begin{cases} y^2-4\sqrt{3}y-\left(22+20\sqrt{3}\right)=0\qquad (5)\\ y^2+4\sqrt{3}y-\left(22-20\sqrt{3}\right)=0\qquad (6) \end{cases}$

$(5)~\Rightarrow~y=2\sqrt{3}\pm\sqrt{34+20\sqrt{3}} ~\Rightarrow~x=4+2\sqrt{3}\pm\sqrt{34+20\sqrt{3}}$

$(6)~\text{has no real solution.}$

$\text{Thus the original equation has two real solutions.}$

$x_1=4+2\sqrt{3}+\sqrt{34+20\sqrt{3}}\simeq15.7491$

$\text{and}$

$x_2=4+2\sqrt{3}-\sqrt{34+20\sqrt{3}}\simeq-0.8208$