Chapter 8.7, Problem 8MRE

To determine

To solve the equation.

Answer to Problem 8MRE

The values of $c\text{ is }-\text{3,}\left(1\pm \sqrt{2}\right)$

Explanation of Solution

Given information:

  $c^3+c^2-7c-3=0;-3$

Calculation:

The equation can be obtained as

  $c^3+c^2-7c-3=0;-3$

  $\begin{array}{l}c=-3\\ \left(c+3\right)=0\end{array}$

Divide the equation $c^3+c^2-7c-3$ by $\left(c+3\right)$

  $\begin{array}{l}c+3\begin{array}{c}\hfill c^2-2c-1\\ \hfill \overline{)c^3+c^2-7c-3}\end{array}\\ \underset{\_}{c^3+3c^2}\\ -\text{2}{c}^2-7c\\ \underset{\_}{-2c^2-6c}\\ -c-3\\ \underset{\_}{-c-3}\\ \text{ 0}\end{array}$

Therefore,

  $\left(c+3\right)\left(c^2-2c-1\right)$

Apply Sridhar Acharya formula to solve the equation

  $\begin{array}{l}\left(c^2-2c-1\right)\\ c=\frac{-\left(-2\right)\pm \sqrt{{\left(-2\right)}^2-4\left(1\right)\left(-1\right)}}{2\left(1\right)}\\ c=\frac{2\pm \sqrt{4+4}}{2}\\ c=\frac{2\pm \sqrt{8}}{2}\\ c=\frac{2\pm 2\sqrt{2}}{2}\\ c=1\pm \sqrt{2}\left[\text{divide numerator and denominator by 2}\right]\end{array}$

Therefore,

  $\begin{array}{l}\left(c+3\right)=0\text{ or }c=1\pm \sqrt{2}\\ c=-3\end{array}$

Hence,

The values of $c\text{ is }-\text{3,}\left(1\pm \sqrt{2}\right)$