How to solve the equation $x^2+2=4\sqrt{x^3+1}$?

Question

From the Leningrad Mathematical Olympiad, 1975:

Solve $x^2+2=4\sqrt{x^3+1}$.

In answer sheet only written $x=4+2\sqrt{3}\pm \sqrt{34+20\sqrt{3}}$.

How to solve this?

Answer 1

I think the "fastest" way to deal with this is surely to square both members, to get

$$x^4 -16x^3 +4x^2 - 12 = 0$$

Now you can treat is as the quartic equation it is, via Ferrari's method for quartic equations.

I will write you down the general procedure for a quartic of the form

$$ax^4 + bx^3 + cx^2 + dx + e = 0$$

And leave you the easy maths behind.

Define the following quantities

$$P = 8ac - 3b^2$$

$$Q = b^3 + 8da - 4abc$$

$$R = c^2 - 3bd + 12 ae$$

$$S = 64a^3e - 16a^2c^2 + 16ab^2c - 16 a^2bd - 3b^4$$

Define other elements

$$U = \frac{P}{8a^2}$$

$$V = \frac{Q}{8a^3}$$

$$Y = 2c^3 - 9bcd + 27b^2e - 27ad^2 - 27ace$$

$$Z = \sqrt[3]{\frac{Y + \sqrt{Y^2 - 4R^2}}{2}}$$

$$W = \frac{1}{2} \sqrt{-\frac{2U}{3} + \frac{1}{3a} \left( Z + \frac{R}{Z}\right) }$$

Final Solutions (note: only two of them will be real, in your case)

$$x_{1,\ 2} = -\frac{b}{4a} - W \pm \sqrt{-4W^2 - 2U + \frac{V}{W}}$$

$$x_{3,\ 4} = -\frac{b}{4a} + W \pm \sqrt{-4W^2 - 2U - \frac{V}{W}}$$

NOTICE In your case $\color{red}{d = 0}$ which simplifies a bit the resolvent.

Answer 2

Let $u = x^2-x+1$, $v = x+1$ and $\lambda = \sqrt{\frac{u}{v}}$, we have

$$\begin{align}u + v & = (x^2-x+1) + (x + 1) = x^2 + 2 \\ &= 4\sqrt{x^3+1} = 4\sqrt{(x+1)(x^2-x+1)}\\ &= 4\sqrt{uv}\end{align}$$ This leads to

$$\lambda^2 + 1 = 4\lambda \implies \lambda = 2 \color{red}{\pm} \sqrt{3}$$ As a result, $$\begin{align} &x^2 - x + 1 = \lambda^2 (x+1) = (7\color{red}{\pm} 4\sqrt{3})(x+1)\\ \iff & x^2 - (8 \color{red}{\pm} 4\sqrt{3})x = (6\color{red}{\pm} 4\sqrt{3})\\ \iff & (x - (4 \color{red}{\pm} 2\sqrt{3}))^2 = (4 \color{red}{\pm} 2\sqrt{3})^2 + (6 \color{red}{\pm} 4\sqrt{3}) = 34 \color{red}{\pm} 20\sqrt{3}\\ \implies & x = 4 \color{red}{\pm} 2\sqrt{3} \color{blue}{\pm} \sqrt{34 \color{red}{\pm} 20\sqrt{3}} \end{align}$$ Notice $34-20\sqrt{3} \approx -0.651 < 0$. For real $x$, the red $\color{red}{\pm}$ sign need to be $+$. This leaves us with

$$x = 4 + 2\sqrt{3} \color{blue}{\pm} \sqrt{34 + 20\sqrt{3}}$$

Answer 3

HINT.-The given answer is root of the quadratic equation $$x^2-2(4+2\sqrt3)x+c=0$$ where $c$ is certain constant. It follows $$x=4+2\sqrt3\pm\sqrt{{(4+2\sqrt3)^2-c}}=4+2\sqrt3\pm\sqrt{34+20\sqrt3}$$ You get so the value of $c$ from which you can verify the solution.