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

Question

Does anyone know how to solve

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

analytically? That is, without using numerical methods to attain an approximate solution.

Answer 1

By the rational root test, this equation doesn't have any rational roots.

So you have to use the general method for solving cubic equations.

Answer 2

Using Cardano's method (and a cas to do the calculations):

we start with

$$x^3+x^2-4=0 \tag{1}$$

and substitute $$x=t-{{1}\over{3}} \tag{2}$$ to get $$t^3-{{t}\over{3}}-{{106}\over{27}}=0 \tag{3}$$

It is always possible to find a substitution $x=t+a$ such that the coefficient of $x^2$ vanishes.

Now we set $$t=v+u \tag{4}$$ and get

$$v^3+{{\left(v+u\right)\,\left(9\,u\,v-1\right)}\over{3}}+u^3-{{106 }\over{27}}=0 \tag{5}$$

We impose the additional condition that $$u\,v={{1}\over{9}} \tag{6}$$

and get

$$v^3+u^3={{106}\over{27}} \tag{7}$$

and

$$u^3\,v^3={{1}\over{729}} \tag{8}$$

So $u^3$ and $v^3$ are the solution of the quadratic equation

$$z^2-{{106\,z}\over{27}}+{{1}\over{729}}=0 \tag{9}$$

This equation has the solutions

$$\left \{-{{6\,\sqrt{78}-53}\over{27}} , {{6\,\sqrt{78}+53 }\over{27}} \right \} \tag{10}$$

$u$ and $v$ are interchangeable, we set

$$u^3=-{{6\,\sqrt{78}-53}\over{27}} \tag{11}$$ $$v^3={{6\,\sqrt{78}+53 }\over{27}} \tag{12}$$

Both values have a real cubic root. We can set

$$u=\sqrt[3]{-{{6\,\sqrt{78}-53}\over{27}}} \tag{13}$$ $$v=\sqrt[3]{{{6\,\sqrt{78}+53 }\over{27}}} \tag{14}$$

because these solution pair satisfies $(1)$.

From this we get $$x=\sqrt[3]{-{{6\,\sqrt{78}-53}\over{27}}} + \sqrt[3]{{{6\,\sqrt{78}+53 }\over{27}}} +1/3 \tag{15}$$

$1$ has three roots in $\mathbb{C}$: $$\left \{ 1 , {{\sqrt{3}\,i}\over{2}}-{{1}\over{2}} , -{{\sqrt{3}\,i}\over{2}}-{{1}\over{2}} \right \} \tag{16}$$ So the two other soltuions we construct from

$$u=(\sqrt[3]{-{{6\,\sqrt{78}-53}\over{27}}}) ({{\sqrt{3}\,i}\over{2}}-{{1}\over{2}}) \tag{17}$$ $$v=(\sqrt[3]{{{6\,\sqrt{78}+53 }\over{27}}}) (-{{\sqrt{3}\,i}\over{2}}-{{1}\over{2}}) \tag{18}$$

and

$$u=(\sqrt[3]{-{{6\,\sqrt{78}-53}\over{27}}}) (-{{\sqrt{3}\,i}\over{2}}-{{1}\over{2}}) \tag{19}$$ $$v=(\sqrt[3]{{{6\,\sqrt{78}+53 }\over{27}}}) ({{\sqrt{3}\,i}\over{2}}-{{1}\over{2}}) \tag{20}$$

in the same way.

Answer 3

There is a general formula to write solutions of any cubic equation. Here is your solution (computed with wolfram alpha): $$ x = \frac{-1+\sqrt[3]{53-6 \sqrt{78}}+\sqrt[3]{53+6 \sqrt{78}}}{3}$$

Answer 4

Perhaps this Wikipedia entry might prove itself useful... (See also the Special cases section below, if necessary).