Solve $x^4+3x+20=0$ by Ferrari's method

Question

Comparing the equation $$x^4+3x+20=0$$
With the equation $$(x^2+\lambda)^2-(mx+n)^2=0$$ we get

$m^2=2\lambda,$

$-2mn=3,$
$n^2=\lambda^2-20$

Now, $4m^2n^2=9\Rightarrow 4(2\lambda)(\lambda^2-20)=9\Rightarrow 8\lambda^3-160\lambda-9=0$.

How can I find easily the values of $\lambda$ from the above equation.

Please suggest me.

Answer 1

The cubic equation is in depressed form (i.e., its quadratic coefficient is 0). To find the first root, you can use Cardano's formula $$\lambda_1=\sqrt[3]{-{q\over 2}+ \sqrt{{q^{2}\over 4}+{p^{3}\over 27}}} +\sqrt[3]{-{q\over 2}- \sqrt{{q^{2}\over 4}+{p^{3}\over 27}}}$$ where $p = -20$ and $q = -\frac{9}{8}$.

Answer 2

You can't find them easily, as the discriminant of the equation is standard form: $$x^3-20x+\frac 98=0$$ is $ 4\cdot (-20)^3+ 27\cdot\Bigl(\dfrac 98\Bigr)<0$.

In such a case, we know the equation has $3$ real toots, but Cardano's formula requires using complex numbers since the square root is that of a negative number, then computing the cubic roots of a complex number.

A trigonometric method is more direct: set $x=A\cos t,\enspace A>0$. The equation becomes $$A^3\cos^3 t -20A\cos t+\frac 98=0.$$ We the choose $A>0$ so that $\dfrac{A^3}4=\dfrac{-20A}3$. Let's call $k$ this common ratio. We get: $$k(4\cos^3t-3\cos t)=k\cos 3t=-\frac 98$$ whence $$3t\equiv \pm\arccos\Bigl(-\frac 9{8k}\Bigr)\mod 2\pi\iff t\equiv\pm\frac 13\arccos\Bigl(-\frac 9{8k}\Bigr)\mod\frac{2\pi}3. $$ These $6$ solutions are equal in pairs, and there are $3$ roots in all, of course.

Answer 3

HINT: Use binom expansion $$(x+y)^3=x^3+3x^2y+3xy^2+y^3$$ Reorder the equation as: $$(x+y)^3-3xy(x+y)-(x^3+y^3)=0$$

$$8\lambda^3-160\lambda-9=0$$
$$\lambda^3-20\lambda-\frac{9}{8}=0$$

Define: $x+y=\lambda$, then solve the equation with 2 unknowns $$3xy=20$$

$$x^3+y^3=\frac{9}{8}$$

Let me know if you cannot get forward from there.