Chapter 8.8, Problem 30WE

To determine

To find the least integral upper bound and greatest integral lower bound for the roots of the equations.

Answer to Problem 30WE

Lower bound $-2$ and upper bound 3.

Explanation of Solution

Given information:

The given equation is $2x^4-6x^2-3x-9=0$ .

Calculation :

For upper bound,

  $\begin{array}{l}3|20-6-3-9\\ 6183690\\ \overline{\text{ 2 6 12 33 }|90}\\ \\ P\left(x\right)=\left(x-3\right)\left(2x^3+6x^2+12x+33\right)+90\end{array}$

There are no roots above 3.

  $\begin{array}{l}P\left(2\right)=2\cdot 16-6\cdot 4-3\cdot 2-9=-7<0\\ P\left(3\right)=90>0\end{array}$

There is a root between 2 and 3, so 3 is the least integral upper bound of the roots.

For lower bound,

  $\begin{array}{l}2|20-6-3-9\\ \text{ 4 }8\text{ 4 14}\\ \overline{\text{ 2 4 2 7 }|5}\\ \\ P\left(-x\right)=2x^4-6x^2+3x-9\\ P\left(-x\right)=\left(x-2\right)\left(2x^3+4x^2+2x+7\right)+5\end{array}$

There are no roots for $P\left(-x\right)$ above 2 and therefore, no roots below $-2$ .

  $\begin{array}{l}P\left(-2\right)=2\cdot 16-6\cdot 4+3\cdot 2-9=5>0\\ P\left(3\right)=2\cdot 1-6\cdot 1+3-9=-10<0\end{array}$

There is root between $-2\text{ and }-1$ , so $-2$ is the greatest integral lower bound for the roots.

Therefore,

Lower bound $-2$ and upper bound 3.