Solving a tricky equation $4x^3-5x^2-5 = 0$

Question

How does one solve $4x^3-5x^2-5 = 0$? I've tried the substitution $y = x - \frac{5}{12}$ but then I ended up with this monster of an equation: $864y^3 - 450y -1205 = 0$. Now I'm stuck. Any help would be greatly appreciated.

Answer 1

If you don't want to use Cardano's formula (even I don't use it, there's a high chance of errors), here's a similar method:

Let $t = \sqrt[3]{a+b}+\sqrt[3]{a-b}$. Then, $$t^3 = 2a+3t\sqrt[3]{a^2-b^2}$$which can be shown simply by cubing both sides. This means that $\sqrt[3]{a+b}+\sqrt[3]{a-b}$ is a (real, assuming $|a|>|b|$ and both $a$ and $b$ were reals) solution of the depressed cubic $$t^3 - 3t\sqrt[3]{a^2-b^2} - 2a = 0$$

We can use this property in the question. Make a substitution $x = 1/y$ to get (note that zero is not a solution of the original equation): $$4(1/y)^3 - 5(1/y)^2 - 5 = 0$$ $$y^3 + y - \frac 45 = 0$$ Here, $a = 2/5$ and hence $b^2 = 133/675$. Thus, $$y = \sqrt[3]{2/5+\sqrt{133/675}}+\sqrt[3]{2/5-\sqrt{133/675}}$$And it can be shown using calculus that this is the only real solution.

Computed $1/y$ using WA to get: $$x = \frac{1}{12}\left(5+\sqrt[3]{1205-60\sqrt{399}}+\sqrt[3]{5(241+12\sqrt{399})}\right)\approx 1.688458990229318741$$ same as the other answer.

Answer 2

Please, do not use Cardano method if you want to avoid poor looking results !

Following the steps given here, you have $$\Delta =-13300 \qquad p=-\frac{25}{48}\qquad q=-\frac{1205}{864}$$ So, one real root.

Using the hyperbolic solution, the real root is $$x=\frac{5}{12} \left(1+2 \cosh \left(\frac{1}{3} \cosh ^{-1}\left(\frac{241}{25}\right)\right)\right)$$

Answer 3

This is not a complete answer but I don't know why but I like to share the value of $x$ that satisfies the given equation. The equation has only one real root and that is $$\frac{1}{12}\left(5+\sqrt[3]{1205-60\sqrt{399}}+\sqrt[3]{5(241+12\sqrt{399})}\right)$$

I wonder if anyone can find it by hand.

Answer 4

This one does not require WA, but requires you to be good at complicated arithmetic:

You have already got $$864y^3 - 450y -1205 = 0 \iff y^3 - \frac{25}{48}y - \frac{1205}{864}=0$$(not the same $y$ in my previous answer). Again, $t = \sqrt[3]{a+b}+\sqrt[3]{a-b}$ is a solution of $$t^3 - 3t\sqrt[3]{a^2-b^2} - 2a = 0$$Comparing, we get $a = \frac{1205}{1728}$, and hence $b^2 = \frac{1436400}{2985984}$. Thus, $b = \frac{60\sqrt{399}}{1728}$. We get: $$y = \sqrt[3]{\frac{1205}{1728}+\frac{60\sqrt{399}}{1728}}+\sqrt[3]{\frac{1205}{1728}-\frac{60\sqrt{399}}{1728}}= \frac1{12}\left(\sqrt[3]{1205+60\sqrt{399}}+\sqrt[3]{1205-60\sqrt{399}}\right)$$ Substituting $x = y+\frac5{12}$ gives the answer: $$\boxed{x = \frac1{12}\left(5+\sqrt[3]{1205+60\sqrt{399}}+\sqrt[3]{1205-60\sqrt{399}}\right)}$$

Answer 5

Slightly different approach very similar to Cardano's method. It uses a series of substitutions.

Given $Ax^3+Bx^2+Cx+D=0$.

Divide by A to get $x^3+Mx^2+Nx+F$

Note that from the symmetry of the cubic, there's a vertical line of symmetry passing directly between the critical points. If you shift the curve with a variable substitution, you can get rid of the quadratic term.

Critical Points: $x= \frac{-2M\pm \sqrt{M^2-12F}}{6}$. Midpoint: $x=-M/3$

Note: Multiply the y values of the two critical points. If this is positive you have one real solution, negative, you have two.

Let $u=x-M/3$ and substitute. You'll get something of the form: $u^3+pu+q=0$.

Now execute another change of variable: $u=(z+G/z)$

Then $u^3+pu+q= z^3+3z^2(G/z)+3z(G/z)^2+(G/z)^3+pz+pG/z+q=0$

Collect like terms and simplify:

$z^3+3Gz+3G/z+G^3/z^3+pz+pG/z+q=z^3+(3G+p)z+(3G+p)/z+G^3/z^3+q=0$

Let $G=-p/3$ to cancel the two terms in the middle, then multiply by $z^3$ to get: $z^6+qz^3+p=0$

Now $z^3=\frac{-q\pm\sqrt{q^2-4p}}{2}$ , $u=z-\frac{p}{3z}$, and $x=u+M/3$