Solving $ x^3 - 3x + \sqrt 2 = 0$

Question

Okay, suppose we have a cubic equation like this: $$x^3 - 3x + \sqrt{2} = 0$$ This is a "depressed cubic" so it should be possible to solve directly.

According to Wikipedia, the rational root test should be able to suggest whether there might be any rational solutions for our equation.

However, this only works for rational solutions, which means that I can't find out whether there's a simple expression for one solution unless that solution is rational. Consequently, if I don't know any of the 3 solutions in advance, then I can't factor it out to get a quadratic.

One way of solving at least a part of the task would be finding a general procedure for taking cube roots. I tried to do that by simply inverting the operations of $a^3 + 3ab^2$ or $3a^2b + b^3$ but I realized that I always had to guess either "a" or "b" (or more precisely, one of its 3 possible values) and then the other one came out. If I made a wrong guess, then the intermediate squares didn't match and I didn't find anything. Therefore, it didn't help me in what I really wanted to get to.

One possible example where it does work is finding the cube root of $387\sqrt{5} - 326\sqrt{7}$: I can divide 387 by 3, which makes 129. I subtract 335 and divide the result by 3, which gives me 28. I divide that by 7 which I took the square root of and I have 4, which is a square of 2 or -2. When I try it the other way round, I see that +2 doesn't work and -2 does, which means that the cube root is: $3\sqrt{5} - 2\sqrt{7}$.

So in this case I really got an expression for the cube root but I had to guess that "a" might be $3\sqrt{5}$ and I got the corresponding "b" then.

Now, to solve $x^3 - 3x + \sqrt{2} = 0$, the rational root test is unusable and therefore I'm supposed to sum two complex cube roots. But it definitely must be possible either to express each of them without using cube roots or to apply a different procedure of finding a solution which can be expressed without them.

Answer 1

$x=\sqrt 2$ is a root by observation. Factor that out and you are left with a quadratic.

How did I guess that? Well the $\sqrt 2$ on its own was matched with odd powers of $x$, so if I wrote $x=y\sqrt 2$ I could see that the $\sqrt 2$ would factor out. Indeed we get $$2\sqrt 2 y^3-3\sqrt 2 y+\sqrt 2=0$$ Then cancel out the $\sqrt 2$ to get $$2y^3-3y+1=0$$ to which you can apply the rational root test etc.

Answer 2

Hint:

You can easily rearrange the statement to:

Plaintext: x^3+\sqrt{2} = 3 x

LaTeX: $x^3+\sqrt{2} = 3 x$

Thus easy to see that

Plaintext: x = sqrt(2), x = -(1+sqrt(3))/sqrt(2), or x = (sqrt(3)-1)/sqrt(2)

LaTeX: $x = \sqrt{2}$, $x = -\frac{1+\sqrt{3}}{\sqrt2}$ or $x = \frac{1+\sqrt{3}}{\sqrt2}$

Solving should prove easy.

Answer 3

Not sure whether this is an answer to your question, but by inspection $\sqrt 2$ is obviously a root. (As Mark Bennet explained I found this by observing that because of which terms the square root was multiplying, a change of variables absorbing the square root into the variable was possible.)

$$x^3-3x+\sqrt 2= (x-\sqrt 2)(x^2 +\sqrt 2 x-1)$$

Answer 4

Substituting $x = \sqrt2$, we get:

$$x^3-3x+\sqrt2 = (x-\sqrt2)(x^2+\sqrt2x-1)$$

Solving the quadratic equation further using the discriminant, we obtain 2 additional roots. Therefore the roots are:

$$x=\sqrt2,\quad x=(-\sqrt2+\sqrt6)/2,\quad x=(-\sqrt2-\sqrt6)/2.$$

Answer 5

By subtracting $\sqrt{2}$ and squaring, you get:

$$x^6 - 6x^4 + 9x^2 = 2$$ $$x^6 - 6x^4 + 9x^2 - 2 = 0$$

Which conveniently has all integer coefficients, although no rational roots. Since all of the exponents are even, substitute $u = x^2$.

$$u^3 - 6u^2 + 9u - 2 = 0$$

And this cubic equation has the rational root $u = 2$. Factoring out $u - 2$ gives:

$$u^2 - 4u + 1 = 0$$

From which the quadratic formula gives:

$$u = 2 \pm \sqrt{3}$$

Thus,

$$u = x^2 \in \{ 2, 2 + \sqrt{3}, 2 - \sqrt{3} \}$$ $$x \in \{ \pm\sqrt{2}, \pm\sqrt{2 + \sqrt{3}}, \pm\sqrt{2 - \sqrt{3}} \}$$

But since squaring introduces extraneous solutions, you have to check which of these solve the original equation $x^3 - 3x + \sqrt 2 = 0 $.

$$\boxed{x \in \left\{ \sqrt{2}, -\sqrt{2 + \sqrt{3}}, \sqrt{2 - \sqrt{3}} \right\}}$$

Answer 6

You can rescue the rational root test by observing that there are only odd powers of $x$. Thus extracting a factor $\sqrt2$ from $x$ allows to cancel the square root from all term. That is, set $x=\sqrt2 y$ to get $$ 2\sqrt2 y^3-3\sqrt2y+\sqrt2 = 0\iff 2y^3-3y+1=0. $$ The rational root method now suggests to try $y=\pm\frac12$ and $y=\pm 1$, and one finds that $y=1$ is indeed a solution.

Answer 7

Another way

$$x^3-3x+\sqrt 2=x^3-2x-x+\sqrt 2= x(x-\sqrt 2)(x+\sqrt 2)-(x-\sqrt 2)=(x-\sqrt 2)(x(x+\sqrt 2)-1)$$

Answer 8

It is usually not possible to solve a depressed cubic by inspection (unless it has rational roots that you can find with the rational root theorem). As shown by others, the given polynomial has been arranged so that trying the root $\sqrt2$ is tempting.

The solution can be found in all cases by assuming that the roots can be expressed as the sum of two terms.

$$x^3+px+q=0,x=u+v\to u^3+(3uv+p)(u+v)+v^3+q=0.$$

By letting $3uv+p=0$, we get the system

$$\begin{cases}uv=-\dfrac p3\to u^3v^3=-\dfrac{p^3}{27}\\u^3+v^3=-q\end{cases}$$

which is of the sum/product type, that we know how to solve.

In your case, $u^3v^3=1$ and $u^3+v^3=-\sqrt2$. Then $(u^3-v^3)^2=(u^3+v^3)^2-4u^3v^3=-2$ so that $u^3,v^3=\dfrac{-\sqrt2\pm i\sqrt2}2$. Possible cubic roots are $\dfrac{1\pm i}{\sqrt2}$, with the sum $\sqrt2$.

Answer 9

Trigonometric substitution trick may help to find the essential root $\sqrt2.$

Let $x=2\cos\theta.$ Then the equation becomes $$\cos3\theta=-\frac1{\sqrt2}$$ whose solution is $\theta=\pm\frac\pi4+\frac{2k\pi}3.$

Hence, $x=2\cos\frac\pi4$, $-2\cos\frac\pi{12}$, $2\sin\frac\pi{12}$.

Answer 10

This answer reached the solution through reverse engineering while avoiding any factorization methods, in particular the use of the Rational Root Theorem.

Let $\sqrt 2=u$. Then we have :

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

We see that, this is a quadratic equation respect to $u$ :

$$x\,u^2-u+(x-x^3)=0$$

So, we have $\Delta_u=4x^4-4x^2+1$ or $\Delta_u=(2x^2-1)^2$.

This tells us,

$$\sqrt 2=u_{1,2}=\frac{1\pm(2x^2-1)}{2x}$$

$$\boxed{\sqrt 2=x \quad \text{or} \quad \sqrt 2=\frac{1-x^2}{x}}$$

Thus, we get:

$$x_1=\sqrt 2$$ and $$x^2+\sqrt{2}\,x-1=0$$

which is a final quadratic equation we need to solve.

Answer 11

Plugging it into the cubic formula we get

$x = \sqrt [3]{\frac {-\sqrt {2}}{2} + i\frac {\sqrt {2}}{2}} + \sqrt [3]{\frac {-\sqrt {2}}{2} - i\frac {\sqrt {2}}{2}}$

$x = \sqrt [3]{e^{\frac {3\pi}{4}i}} + \sqrt [3]{e^{-\frac {3\pi}{4}i}} $

$e^{\frac {\pi}{4}i} + e^{-\frac {\pi}{4}i}\\ 2\cos (\frac {\pi}{4}) = \sqrt 2$

Other solutions are:

$2\cos (\frac {\pi}{4} + \frac {2\pi}{3}) = 2\cos (\frac {11\pi}{4}) = -\frac {\sqrt 6 + \sqrt 2}{2}\\ 2\cos (\frac {\pi}{4} - \frac {2\pi}{3}) = 2\cos (-\frac {5\pi}{12}) = \frac {\sqrt {6}-\sqrt 2}{2}$

I just thought of an approach that I think I like better.

Note that $\cos 3\theta = 4\cos^3\theta - 3\cos \theta$

let $x=2\cos \theta$

$8\cos^3\theta - 6\cos\theta + \sqrt 2 =0$

$4\cos^3 \theta -3\cos\theta = -\frac {\sqrt 2}{2}$

$\cos 3\theta = -\frac {\sqrt 2}{2}\\ 3\theta = \frac {3\pi}{4} + 2n\pi\\ \cos \theta = \cos (\frac {\pi}{4} + \frac {2n\pi}{3})\\ x = 2\cos\theta = 2 \cos (\frac {\pi}{4} + \frac {2n\pi}{3})$

Answer 12

If we multiply it by $\sqrt{8}$ we get $$ (\sqrt{2}x)^3 - 6(\sqrt{2}x) + 4 = 0$$

Now let $t=x\sqrt{2}$ then we have $$t^3-6t+4=0$$ Now try rational root theorem.