(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.
thanks in advance
(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.
thanks in advance
First, that's not a polynomial equation. Second there is a missing "(" on the left. Is it supposed to be $\displaystyle (x^2+ 2)/2$?
And it is hard to read- that is $\displaystyle x^3+ 1$ on the right, not $\displaystyle x^2+ 1$ as emkarov seemed to think.
The first thing you should do is make it a polynomial equation by squaring both sides.
$\displaystyle \frac{x^4+ 4x^2+ 4}{4}= 4(x^3+ 1)$
$\displaystyle x^4+ 4x^2+ 4= 16x^3+ 16$
$\displaystyle x^4- 16x^3+ 4x^2- 12= 0$.
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.
$\displaystyle \text{Currently I cannot find a substitution method to solve this problem. However, it can be solved by brute force!}$
$\displaystyle x^2+2=4\sqrt{x^3+1}~\Leftrightarrow~x^4-16x^3+4x^2-12=0\qquad (1)$
$\displaystyle \text{There is no real rational solution to (1), therefore we have to solve the \emph{quartic equation} using \textbf{Ferrari's Method}}. $
$\displaystyle \text{Let}~ x=y-\frac{b}{4a}~\Rightarrow~x=y+4.~\text{This substitution will transform}~(1)~\text{into:}$
$\displaystyle y^4-92y^2-480y-716=0~\Leftrightarrow~y^4=92y^2+480y+716$
$\displaystyle \text{Add}~2zy^2+z^2~\text{to both sides of the equation, for some}~z\in R.~\text{The added terms will make the left side of the equation a perfect square.}$
$\displaystyle \left(y^2+z\right)^2=(2z+92)y^2+480y+716+z^2\qquad (2)$
$\displaystyle \text{The right side is a perfect square if its discriminant is equal to 0, since it is a quadratic equation in terms of}~y.$
$\displaystyle D=230400-4(2z+92)(z^2+716)=0~\Leftrightarrow~28800-(z+46)(z^2+716)=0 $
$\displaystyle \Leftrightarrow~z^3+46z^3+716z+4136=0\qquad(3) $
$\displaystyle \text{Let}~z=u-\frac{b}{3a}~\Leftrightarrow~z=u-\frac{46}{3}\qquad (4) $
$\displaystyle (3)~\Leftrightarrow~u^3+\frac{32}{3}u+\frac{9920}{ 27}~\Leftrightarrow~27u^3+288u+9920=0$
$\displaystyle \Leftrightarrow~(3u+20)(9u^2-60u+496)=0~\Rightarrow~u=-\frac{20}{3}~\Rightarrow~z=-22$
$\displaystyle (2)~\Leftrightarrow~\left(y^2-22\right)^2=48y^2+480y+1200=48(y+5)^2$
$\displaystyle \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} $
$\displaystyle (5)~\Rightarrow~y=2\sqrt{3}\pm\sqrt{34+20\sqrt{3}} ~\Rightarrow~x=4+2\sqrt{3}\pm\sqrt{34+20\sqrt{3}} $
$\displaystyle (6)~\text{has no real solution.}$
$\displaystyle \text{Thus the original equation has two real solutions.}$
$\displaystyle x_1=4+2\sqrt{3}+\sqrt{34+20\sqrt{3}}\simeq15.7491$
$\displaystyle \text{and} $
$\displaystyle x_2=4+2\sqrt{3}-\sqrt{34+20\sqrt{3}}\simeq-0.8208$