Answer to Question #177018 in Quantitative Methods for Mary

Question

Given the quantic equation below, solve it to find the values of x.

 

 2𝑥5 − 6𝑥3 − 4𝑥2 + 2𝑥 + 4 = 0 

Answer 1

Simplify the equation. Divide left and right by the common factor 2:

$$x^5-3x^3-2x^2+x+2=0$$

Use Descartes’ Rule of Signs to find how many positive and how many negative real zeroes the equation has.

Let $p(x) = x^5-3x^3+2x^2+x+2$ . p(x) has 2 variations in sign, there must be either two positive roots, or no positive roots.

$$p(-x) = -x^5+3x^3-2x^2-x+2=0$$

p(−x) has three variation in sign, and therefore the original p(x) has three negative roots or one negative root.

Use Rational Root Test to find rational roots. The factors of the constant term (2) are 1, 2, -1,-2. Check whether these numbers are roots of the equation:

$$p(1)=1^5-3\cdot1^3-2\cdot 1^2+1+2=1-3-2+1+2=-1\mathrlap{\,/}{=}0$$ $$p(2)=2^5-3\cdot2^3-2\cdot 2^2+2+2=32-24-8+2+2=4\mathrlap{\,/}{=}0$$ $$p(-1)=-1^5-3\cdot(-1)^3-2\cdot (-1)^2-1+2=-1+3-2-1+2=1\mathrlap{\,/}{=}0$$ $$p(-2)=(-2)^5-3\cdot(-2)^3+2\cdot (-2)^2-2+2=-32+24-8-2+2=-16$$

Thus, the equation has no rational roots.

Since p(1) and p(2) have opposite signs, there is at least one root between those numbers. Also, there are roots between -2 and -1, -1 and 1.

1) Find the approximate value of root between 1 and 2, using Newton's method.

$$p'(x)=5x^4-9x^2-4x+1$$

Let x₀ = 2. Then $x_1 =x_0 -\frac{f(x_0)}{f'(x_0)}=2-\frac{4}{37}\approx1.9$

$x_2=x_1 -\frac{f(x_1)}{f'(x_1)}\approx1.9-\frac{0.86}{26}\approx1.87$

2) Find the approximate value of root between 1 and -1,

Let x₀ = 1. Then $x_1 =x_0 -\frac{f(x_0)}{f'(x_0)}=1-\frac{-1}{-7}\approx0.9$

$x_2=x_1 -\frac{f(x_1)}{f'(x_1)}\approx0.9-\frac{-0.31}{-6.6}\approx0.85$

3) Find the approximate value of root between -2 and -1,

Let x₀ =-2. Then $x_1 =x_0 -\frac{f(x_0)}{f'(x_0)}=-2-\frac{-16}{61}\approx-1.7$

$x_2=x_1 -\frac{f(x_1)}{f'(x_1)}\approx-1.7-\frac{-4.93}{23.55}\approx-1.5$

$x_3=x_2-\frac{f(x_2)}{f'(x_2)}\approx-1.5-\frac{-1.46}{12.06}\approx-1.4$

$x_4=x_3-\frac{f(x_3)}{f'(x_3)}\approx-1.4-\frac{-0.47}{8.17}\approx-1.34$

Answer: Approximate values of roots are: -1.34, 0.85, 1.87