Answer to Question #103604 in Algebra for ea

Question

Find the roots of 2x^4 + 9x^3 -9x^2 - 46x +24 by Ferrari’s method.

Answer 1

To solve this equation by Ferrari’s method, first divide each term of the equation by the leading coefficient (2). Obtain

$x^4 + \frac{9}{2}x^3 - \frac{9}{2}x^2 - 23x + 12 = 0$

Denote the coefficients of the obtained equation as $a=\frac{9}{2}, b=-\frac{9}{2}, c=-23, d=12$.

Next, you should determine any root of the following equation.

$y^{3}-by^{2}+(ac-4d)y-a^{2}d+4bd-c^{2}=0$

Substituting $a,b,c,d$, obtain

$y^3 + \frac{9}{2}y^2 -\frac{300}{2}y-988=0$

By substitution method determine that the root is $y_0 = -8$.

Then the needed roots is the roots of the next equations.

$x^{2}+{\frac{a}{2}}x+{\frac {y_{0}}{2}}=\pm {\sqrt {\left({\frac{a^{2}}{4}}-b+y_{0}\right)x^{2}+\left({\frac{a}{2}}y_{0}-c\right)x+{\frac{y_{0}^{2}}{4}}-d}}$

Substituting $a,b,c,d, \text{and } y_0$, obtain

$x^{2}+{\frac{9}{24}}x - 4=\pm \sqrt{\frac{25}{16}x^2 + 5x + 4}$

The square-root expressions is the perfect square.

$x^{2}+{\frac{9}{24}}x - 4=\pm \sqrt{\left(\frac{5}{4}x + 2\right)^2}$

Thus, obtain two equation. Solve the first equation.

$x^{2}+{\frac{9}{24}}x - 4=\frac{5}{x}x+2 \\ \\ x^{2}+x-6=0$

The roots are $-3$ and $2$.

Solve the second equation.

$x^{2}+{\frac{9}{24}}x - 4=-\frac{5}{x}x-2 \\ \\ x^{2}+\frac{7}{2}x-2=0$

The roots are $-4$ and $\frac{1}{2}$.

Therefore, the roots of the original equation are $-4, -3, \frac{1}{2}, \text{and } 2$.