Chapter 8.7, Problem 9WE

To determine

To find the rational roots of the equation.

Answer to Problem 9WE

The rational roots are $\frac{3}{2},2\pm \sqrt{2}$

Explanation of Solution

Given information:

  $2x^3-11x^2+16x-6=0$

Calculation:

The possible rational roots can be obtained as

  $2x^3-11x^2+16x-6=0$

First find h and k

  $\begin{array}{l}h=1,2,3\text{ or 6 }\left[\begin{array}{l}h=\text{factor of the constant term of the polynomial}\\ \text{since, 6 is divisible by 1,2,3 or 6}\end{array}\right]\\ k=1\text{ or }2\left[\begin{array}{l}\text{k= factor of the cofficient of the highest-degree term}\\ \text{since, 2 is divisible by 1 and 2}\end{array}\right]\end{array}$

  $$ All possible rational roots are in the form of $\pm \frac{h}{k}$

Therefore,

  $\begin{array}{l}\pm \frac{1}{1}=\pm 1\text{ or }\pm \frac{2}{1}=\pm 2\text{ or }\pm \frac{3}{1}=\pm 3\text{ or }\pm \frac{6}{1}=\pm 6\\ \pm \frac{1}{2}=\pm \frac{1}{2}\text{ or }\pm \frac{2}{2}=\pm 1\text{ or }\pm \frac{3}{2}=\pm \frac{3}{2}\text{ or }\pm \frac{6}{2}=\pm 3\end{array}$

  $\begin{array}{l}\overline{)1}\text{ 2 }-11\text{ 16 }-6\\ \underset{\_}{\text{ 2 }-9\text{ 7}}\\ \text{ 2 }-9\text{ 7 }⋮\text{ 1}\leftarrow \text{ 1 is not a root}\\ -\overline{)1}\text{ 2 }-11\text{ 16 }-6\\ \underset{\_}{-\text{ 2 13 }-29}\\ \text{ 2 }-13\text{ 29 }⋮-35\leftarrow -\text{1 is not a root}\\ \overline{)2}\text{ 2 }-11\text{ 16 }-6\\ \underset{\_}{\text{ 4 }-14\text{ 4}}\\ \text{ 2 }-7\text{ 2 }⋮-2\leftarrow \text{ 2 is not a root}\\ \overline{)3}\text{ 2 }-11\text{ 16 }-6\\ \underset{\_}{\text{ 6 }-15\text{ 3}}\\ \text{ 2 }-5\text{ 1 }⋮-3\leftarrow \text{ 3 is not a root}\\ \overline{)6}\text{ 2 }-11\text{ 16 }-6\\ \underset{\_}{\text{ 12 6 132}}\\ \text{ 2 1 22 }⋮\text{ 126}\leftarrow \text{ 4 is not a root}\\ \overline{)\frac{1}{2}}\text{ 2 }-11\text{ 16 }-6\\ \underset{\_}{\text{ 1 }-5\frac{11}{2}}\\ \text{ 2 }-10\text{ 11 }⋮-\frac{1}{2}\leftarrow \frac{1}{2}\text{ is not a root}\\ \overline{)\frac{3}{2}}\text{ 2 }-11\text{ 16 }-6\\ \underset{\_}{\text{ 3 }-\text{12 6}}\\ \text{ 2 }-8\text{ 4 }⋮\text{ 0}\leftarrow \frac{3}{2}\text{ is a root}\end{array}$

The depressed equation is $2x^2-8x+4=0$

Apply Sridhar Acharya formula to solve the above equation

  $\begin{array}{l}2x^2-8x+4=0\\ x=\frac{-\left(-8\right)\pm \sqrt{{\left(-8\right)}^2-4\left(2\right)\left(4\right)}}{2\left(2\right)}\\ x=\frac{8\pm \sqrt{64-32}}{4}\\ x=\frac{8\pm \sqrt{32}}{4}\\ x=\frac{8\pm \sqrt{4\times 4\times 2}}{4}\\ x=\frac{8\pm 4\sqrt{2}}{4}\left[\text{since,}\sqrt{4\times 4}=4\right]\\ x=2\pm \sqrt{2}\left[\text{divide numerator and denominator by 4}\right]\end{array}$

Hence,

The rational roots are $\frac{3}{2},2\pm \sqrt{2}$