# How do you solve 4x^3 -17x^2 -4 = 0 ?

May 5, 2018

$\text{There is a method to solve a cubic equation in general by hand}$ $\text{(and calculator) on paper. It is a method based on the substi-}$
$\text{tution of Vieta.}$

$\text{Dividing by the first coefficient yields :}$
${x}^{3} - \left(\frac{17}{4}\right) {x}^{2} - 1 = 0$
$\text{Substituting "x=y+p" in "x^3+ax^2+bx+c" yields :}$
${y}^{3} + \left(3 p + a\right) {y}^{2} + \left(3 {p}^{2} + 2 a p + b\right) y + {p}^{3} + a {p}^{2} + b p + c = 0$
$\text{if we take "3p+a=0 => p=-a/3", the first coefficient becomes}$ $\text{zero, and we get :}$
${y}^{3} - \left(\frac{289}{48}\right) y - \left(\frac{5777}{864}\right) = 0$
$\text{(with p = 17/12)}$
$\text{Substituting y=qz in "y^3 + b y + c = 0", yields :}$
${z}^{3} + b \frac{z}{q} ^ 2 + \frac{c}{q} ^ 3 = 0$
$\text{if we take "q = sqrt(|b|/3)", the coefficient of z becomes 3 or -3,}$
$\text{and we get :}$
$\text{(here q = 1.41666667)}$
${z}^{3} - 3 z - 2.35171993 = 0$
$\text{Substituting z = t + 1/t, yields :}$
${t}^{3} + \frac{1}{t} ^ 3 - 2.35171993 = 0$
$\text{Substituting "u = t^3", yields the quadratic equation :}$
${u}^{2} - 2.35171993 u + 1 = 0$
$\text{A root of this quadratic equation is u=1.79444436.}$

$\text{Substituting the variables back, yields :}$
$t = \sqrt{u} = 1.21518761 .$
$\implies z = 2.03810581 .$
$\implies y = 2.88731656 .$
$\implies x = 4.30398323 .$
$\text{The other roots can be found by dividing and solving the}$ $\text{remaining quadratic equation.}$
$\text{The other roots are complex : "-0.02699161 pm 0.48126330 i.}$