Chapter 8.8, Problem 17WE

To determine

To calculate : the real roots of the given equation $P(x)=0$ and approximate it to the nearest tenth.

Answer to Problem 17WE

The roots of the equation are $-0.3,-1.5,\text{ and }1.9$

Explanation of Solution

Given information :

The polynomial equation:

  $P(x)=x^3-3x-1$

Formula used :

Intermediate-Value Theorem:

If P is a polynomial function with real coefficients, and m is any number between P (a) and P ( b ), then there is at least one number c between a and b for which P ( c ) = m .

Calculation :

As per the given problem,

Consider the given equation:

  $P(x)=x^3-3x-1$

By Descartes’s rule $P(x)=0$ has one positive root and two negative roots. Make a table using negative and positive values of x.

The table of values at the left below shows that $P(-2)<0$ and $P(-1)>0$ , and again $P(0)<0$ . Also $P(1)<0$ and $P(2)>0$ . Therefore, one negative root lies between $-2$ and $-1$ , and the other lies between $-1$ and $0$ , and the positive root lies between $1$ and $2$ .

  

The table of values at the right above, shows that the positive root is between $1.8$ and $1.9$ Since $P(1.9)$ is closer to $0$ than $P(1.8)$ , it can be assumed that root is closer to $1.9$ . Therefore, to the nearest tenth, the positive root of P is $1.9$

The table of values at the left below, shows that one of the negative root is between $-1.6$ and $-1.5$ Since $P(-1.5)$ is closer to $0$ than $P(-1.6)$ , it can be assumed that root is closer to $-1.5$ . Therefore, to the nearest tenth, one of the negative roots of P is $-1.5$

The table of values at the right below, shows that the other negative root is between $-0.4$ and $-0.3$ Since $P(-0.3)$ is closer to $0$ than $P(-0.4)$ , it can be assumed that root is closer to $-0.3$ . Therefore, to the nearest tenth, the other negative root of P is $-0.3$

  

The roots of P are $-0.3,-1.5,\text{ and }1.9$