Chapter 4.5, Problem 9E

To determine

To solve: the given equation.

Answer to Problem 9E

The solution of the given equation is $c=0,3,\sqrt{6}$ and $c=-\sqrt{6}$ .

Explanation of Solution

Given information:

A quartic equation is given as

  $2c^4-6c^3=12c^2-36c$

Calculation:

Consider the given equation.

  $2c^4-6c^3=12c^2-36c$

Simplify the above equation as shown:

  $\begin{array}{l}2c^4-6c^3-12c^2+36c=12c^2-36c-12c^2+36c\\ 2c^4-6c^3-12c^2+36c=0\end{array}$

Take common $2c$ from each term of left side of the equation.

  $2c\left(c^3-3c^2-6c+18\right)=0$

Now, factorize left side of $2c\left(c^3-3c^2-6c+18\right)=0$ as shown:

  $\begin{array}{c}2c\left(c^3-3c^2-6c+18\right)=0\\ 2c\left[c^2\left(c-3\right)-6\left(c-3\right)\right]=0\\ 2c\left(c-3\right)\left(c^2-6\right)=0\end{array}$

Solution can be found by equating each factor with 0.

  $c=0$

And

  $\begin{array}{c}c-3=0\\ c=3\end{array}$

Again, factorize and solve $c^2-6=0$ as shown:

  $\begin{array}{l}{c}^2-6=0\\ {\left(c\right)}^2-{\left(\sqrt{6}\right)}^2=0\\ \left(c+\sqrt{6}\right)\left(c-\sqrt{6}\right)=0\end{array}$

Further,

  $\begin{array}{c}c+\sqrt{6}=0\\ c=-\sqrt{6}\end{array}$ and $\begin{array}{c}c-\sqrt{6}=0\\ c=\sqrt{6}\end{array}$

Therefore, the solution of the given equation is $c=0,3,\sqrt{6}$ and $c=-\sqrt{6}$ .