Question

1.a-Solve 9x^4-18x^3-31x^2 8x 12=0 by ferrari's method. 2.b-solve the equeation in(a) given that the sum of two of its roots is zero

Answer 1

Answer on Question #46767 – Math – Algorithms | Quantitative Methods

1.a-Solve $9x^4 - 18x^3 - 31x^2 + 8x + 12 = 0$ by Ferrari's method.

Solution:

Solving bi-quadratic polynomials can be done using Ferrari's method, which transforms a bi-quadratic polynomial into a depressed bi-quadratic which has no $x^3$ term.

By substituting $x = y - \frac{b}{4a}$ we arrive at the above equation $y^4 + py^2 + qy + r = 0$

Where $p = \frac{8ac - 3b^2}{8a^2}$, $q = \frac{8a^2d + b^3 - 4abc}{8a^3}$, $r = \frac{16ab^2c - 64a^2bd - 3b^4 + 256a^3e}{256a^4}$.

If $q \neq 0$ then we have to solve the auxiliary cubic equation.

$$z^3 + p z^2 + \frac{p^2 - 4r}{4} z - \frac{q^2}{8} = 0$$

Start to find the value of $q$. In our case we have $a = 9$, $b = -18$, $c = -31$, $d = 8$, $e = 12$.

Substitute the given values into the formula for $q$ noted above.

$$\begin{array}{l} q = \frac{8a^2d + b^3 - 4abc}{8a^3} = \frac{8 \cdot (9)^2 \cdot 8 + (-18)^3 - 4(9)(-18)(-31)}{8(9)^3} \\ = \frac{5184 - (-5832) - 20088}{5832} = -1.55556 \end{array}$$

If $q \neq 0$ then this equation is always a positive root, which we denote $z_0$. Then the roots of the original equation can be obtained from the formulas.

$$z^3 + p z^2 + \frac{p^2 - 4r}{4} z - \frac{q^2}{8}$$

Find the value of $p$.

$$p = \frac{8ac - 3b^2}{8a^2} = \frac{8(9)(-31) - 3(-18)^2}{8(9)^2} = \frac{-2232 - 972}{648} = -4.9444$$

Calculate the value of $r$.

$$\begin{array}{l} r = \frac{16ab^2c - 64a^2bd - 3b^4 + 256a^3e}{256a^4} \\ = \frac{16(9)(-18)^2(-31) - 64(9)^2(-18)(8) - 3(-18)^4 + 256(9)^3(12)}{256(9)^4} \\ = 0.729167 \\ \end{array}$$ $$r = 0.729167$$

Then we solve the following equation.

$$z ^ {3} + (- 4. 9 4 4) z ^ {2} + \frac {(- 4 . 9 4 4) ^ {2} - 4 (0 . 7 2 9)}{4} z - \frac {(- 1 . 5 5 6) ^ {2}}{8} = 0$$

Simplify to find the value of $z$ .

$$z ^ {3} - 4. 9 4 4 z ^ {2} + 5. 3 8 2 z - 0. 3 0 3 = 0$$

As a result of solution we obtained the following roots.

$$z _ {1} \approx 0. 0 5 9 5$$ $$z _ {2} \approx 1. 5 0 7 7$$ $$z _ {3} \approx 3. 3 7 6 7 2$$

Now we can find the root of the original equation.

$$y _ {1} = \frac {\sqrt {2 z _ {0}} - \sqrt {2 z _ {0} - 4 (\frac {p}{2} + z _ {0} + \frac {q}{2 \sqrt {2 z _ {0}}})}}{2}$$ $$y _ {2} = \frac {\sqrt {2 z _ {0}} + \sqrt {2 z _ {0} - 4 (\frac {p}{2} + z _ {0} + \frac {q}{2 \sqrt {2 z _ {0}}})}}{2}$$ $$y _ {3} = \frac {- \sqrt {2 z _ {0}} - \sqrt {2 z _ {0} - 4 (\frac {p}{2} + z _ {0} - \frac {q}{2 \sqrt {2 z _ {0}}})}}{2}$$ $$y _ {4} = \frac {- \sqrt {2 z _ {0}} + \sqrt {2 z _ {0} - 4 (\frac {p}{2} + z _ {0} - \frac {q}{2 \sqrt {2 z _ {0}}})}}{2}$$

Of the three roots of the equation, we choose the root equal to $z_{3} \approx 3.37672$ .

$$y _ {1} = 0. 2 5 9$$ $$y _ {2} = 2. 3 4 0$$ $$y _ {3} = - 2. 3 4 0$$ $$y _ {4} = - 0. 2 5 9$$ $$x _ {1} = y - \frac {b}{4 a} = 2. 3 4 0 - \frac {- 1 8}{4 (9)} = 2. 3 4 0 + 0. 5 = 2. 8 4$$ $$x _ {2} = y - \frac {b}{4 a} = 0. 2 5 9 - \frac {- 1 8}{4 (9)} = 0. 2 5 9 + 0. 5 = 0. 7 5 9$$ $$x _ {3} = - 2. 3 4 0 + 0. 5 = - 1. 8 4$$ $$x _ {4} = - 0. 2 5 9 + 0. 5 = 0. 2 4 1$$

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