Irrational equation $\left(x^{3}-3x+1\right)\sqrt{x^{2}-1}+x^{4}-3x^{2}+x+1=0$

Question

I saw the problem from one math competition: $$\left(x^{3}-3x+1\right)\sqrt{x^{2}-1}+x^{4}-3x^{2}+x+1=0$$.

I tried to solve it this way: \begin{align*} & \left(x^{3}-3x+1\right)\sqrt{x^{2}-1}+x^{4}-3x^{2}+x+1=0\\ \Leftrightarrow\,\, & \left(x^{3}-3x+1\right)\sqrt{x^{2}-1}+x\left(x^{3}-3x+1\right)+1=0\\ \Leftrightarrow\,\, & \left(x^{3}-3x+1\right)\left(\sqrt{x^{2}-1}+x\right)=-1 \end{align*}

Now I tried to use substitutions like $a=x^3-3x$ and $b=\sqrt{x^{2}-1}$ but I cannot get anything. I know that solutions are $x=1$ and $x=\pm\sqrt{2}$.

Answer 1

Of course, $x= \pm \sqrt{2}$ are two roots of this equation. Assume $x^2 \ne 2$, we have: \begin{align*} &(x^{3} - 3x + 1)(\sqrt{x^{2} - 1} + x) = -1 &\Longleftrightarrow & x^{3} - 3x + 1 = \sqrt{x^{2} - 1}-x \\\\ \Longleftrightarrow & x^{3} - 2x + 1 = \sqrt{x^{2} - 1} & \Longleftrightarrow & x^{3} - 2x = \sqrt{x^{2} - 1}-1\\\\ \Longleftrightarrow & (x^{2} - 2)x = \frac{x^2-2}{\sqrt{x^{2} - 1}+1} & \Longleftrightarrow & x = \frac{1}{\sqrt{x^{2} - 1}+1} \end{align*} Clearly, the only root for the last equation is $1$ because $LHS \ge 1 \ge RHS$.
So, the initial equation has exactly 3 roots $\{ 1, \sqrt{2},-\sqrt{2}\}$.

Answer 2

$$(x^3-3x+1)\sqrt{x^2-1}+x^4-3x^2+x+1=0$$ $$(x^3-3x+1)\sqrt{x^2-1}=-(x^4-3x^2+x+1)$$ $$(x^3-3x+1)^2(x^2-1)=(x^4-3x^2+x+1)^2$$

If you expand everything, and then subtract the two sides of the equation, you get

$$x^6-4x^4+2x^3+3x^2-4x+2 = 0$$

By the Rational Root Theorem, potential rational roots are $\pm 1$ and $\pm 2$, of which it turns out that $x = 1$ is the only one that works. Divide the polynomial by $x - 1$.

$$x^5+x^4-3x^3-x^2+2x-2=0$$

Unfortunately, there's no Quintic Formula. And no rational roots. But maybe we can find a quadratic factor somehow.

$$(x^3+ax^2+bx+c)(x^2+dx+e)=0$$ $$x^5+(a+d)x^4+(b+ad+e)x^3+(c+bd+ae)x^2+(cd+be)x+ce=0$$

It's a bit tedious, but we can match up the coefficients and solve the system of 5 equations for 5 variables, ultimately getting

$$(x^3+x^2-x+1)(x^2-2)=0$$

The quadratic factor obviously gives the roots $x = \pm \sqrt{2}$.

The cubic, AFAICT, has no simple solution, so I solved it numerically to get $x \approx -1.8392867552141612$. However, this turns out to be an extraneous solution that does not solve the original equation. The other two roots of the cubic are complex numbers, so I assume you're not interested in them.

Therefore, $x \in \lbrace 1, \sqrt{2}, -\sqrt{2} \rbrace$.

Answer 3

A complete solution, building on your last identity.

Multiply your last equation by $x-\sqrt{x^2-1}\neq0$ to obtain:

$$\begin{align}x^3-3x+1&=\sqrt{x^2-1}-x\\x^3-2x+1&=\sqrt{x^2-1}\\x(x^2-2)+1&=\sqrt{(x^2-2)+1}\end{align}$$

$1$ is a solution by inspection of the second equation, and $\pm\sqrt{2}$ are solutions by inspection of the third equation, which was hopefully well motivated from a natural rearrangement.

Now taking $x$ not equal to one of the already determined solutions, square both sides, potentially introducing extraneous solutions: $$\begin{align}(x^3-2x+1)^2&=x^2-1\\x^6-4x^4+2x^3+3x^2-4x+2&=0\end{align}$$

We know that we can divide by $(x-1)$ and also by $x^2-2$, so we reduce to: $$(x-1)(x^5+x^4-3x^3-x^2+2x-2)=0\\(x^2-2)(x^3+x^2-x+1)=0$$

Now we know any solution of our original problem is either a solution already found, or it is a (real) root of $x^3+x^2-x+1=0$.

For such a root, the quantity $x^3-2x+1$ is equal to $-(x^2+x)$. Suppose $x\gt0$: then $0\gt0+1-x,x\gt1$ but this implies $0\gt1+1+1-x$ and $x\gt3$, so on, we have a contradiction in the magnitude of $x$. Moreover $x\gt3$ is impossible since Cauchy’s root bound ensures $|x|\le2$. Lets check the quantity $x^2+x$ - it is negative iff. $x\lt-x^2$, which forces that $-1\lt x$ by a comparison of magnitude. But then $0\gt-1+x^2-x+1=x^2-x$ and $x\gt x^2$, a contradiction. Thus the quantity $x^2+x$ is positive for all roots of the equation $x^3+x^2-x+1$. Were this root a solution of our desired equation, then $-(x^2+x)=\sqrt{x^2-1}$, but this is an impossibility as the left hand side is negative and the right hand side is positive. Thus this other root is extraneous and we can ignore it, and $1,\pm\sqrt{2}$ are the only solutions.

Answer 4

HINT

Based on the suggestion of @JohnOmielan in the comments, assuming that $|x| > 1$, it yields that \begin{align*} (x^{3} - 3x + 1)(\sqrt{x^{2} - 1} + x) = -1 & \Longleftrightarrow x^{3} - 3x + 1 = \sqrt{x^{2} - 1} - x\\\\ & \Longleftrightarrow x^{3} - 2x + 1 = \sqrt{x^{2} - 1}\\\\ & \Longleftrightarrow (x^{3} - 2x + 1)^{2} = x^{2} - 1\\\\ & \Longleftrightarrow x^{6} - 4x^{4} + 2x^{3} + 4x^{2} - 4x + 1 = x^{2} - 1\\\\ & \Longleftrightarrow x^{6} - 4x^{4} + 2x^{3} + 3x^{2} - 4x + 2 = 0 \end{align*}

Since the sum of the coefficients equals zero, the value $x = 1$ is a root.

Taking advantage of such fact, can you proceed from here?

Answer 5

Write $P\equiv \sqrt{x^2-1}$ and $Q\equiv x^3-3x+1$ then the first equation can be written as $$ Q(x+P)+1=0 \tag{1} $$ It's clear that $|x|\geq 1$ otherwise there is no solution for $P$. By inspection, $x=1$ is one of the solutions. Assuming $|x|>1$, multiply $(1)$ by $(P-x)$: $$ Q(P^2-x^2)+P-x=0 $$ Since $P^2-x^2=-1$, $$ P=Q+x=x^3-2x+1 $$ Define $y=x^2-2$ then this becomes $$ P=yx+1. $$ Squaring, $$ P^2\equiv y+1=(yx+1)^2 $$ and so $$ y+1=(yx+1)^2 $$ If $x>1$, you can easily convince yourself that the only way for this to be true is if $y=0$, hence $x=\pm \sqrt2$.

However, there is a flaw that I have not yet convinced myself that if $x<-1$ there is no other obvious solution than $x=-\sqrt2$. Sorry to leave a slightly incomplete answer.

Answer 6

Putting another answer here, a simpler one. We start by multiplying both sides of the equation by $x\sqrt{x^2-1}$. This gives $$(x^4-3x^2+x+1)\sqrt{x^2-1}+(x^5-4x^3+x^2+3x-1)=0$$

Subtracting this from the original equation, we get directly: $$\sqrt{x^2-1}=x^3-2x+1$$ We can see that $x=1$ is a solution and other solution must be for $|x|>1$.

We now look at $f(x)=x^3-2x+1-\sqrt{x^2-1}$. Its derivative is $f'(x)=3x^2-2-\frac x{\sqrt{x^2-1}}$.

For $x<-1$, $f'(x)$ always positive, so there is only one solution, which is $x=-\sqrt2$.

For $x>1$, we check that $f''(x)$ is always positive. So we can have only 2 solutions. We already have $x=1$ so the only one left is $x=\sqrt2$.