Chapter 5.7, Problem 42E

To determine

To state: The number of roots of the equation $4x^3-4x^2+13x-6=0$ . Then solve the equation.

Answer to Problem 42E

  $\begin{array}{l}{x}_1=\frac{1}{2}\\ x_{2,3}=\frac{1\pm i\sqrt{47}}{4}\end{array}$

Explanation of Solution

Given information: $4x^3-4x^2+13x-6=0$

Calculation:

The equation is of the third order,

Let’s factor out 4x from the first three terms,

We are trying to express the first three terms as a square of a binomial.

So we expect 3 solutions

(Some real, some maybe complex)

  $\begin{array}{l}4x^3-4x^2+13x-6=0\\ 4x\left(x^2-x+\frac{1}{4}\right)-4x\cdot \frac{1}{4}+13x-6=0\\ 4x{\left(x-\frac{1}{2}\right)}^2+12\left(x-\frac{1}{2}\right)=0\\ \left(x-\frac{1}{2}\right)\cdot \left[4x\left(x-\frac{1}{2}\right)+12\right]=0\\ \left(x-\frac{1}{2}\right)\cdot \left(4x^2-2x+12\right)=0\\ \left(x-\frac{1}{2}\right)\cdot 2\cdot \left(2x^2-x+6\right)=0\end{array}$

Luckily enough, the other terms fall into place and we have common parentheses $\left(x-\frac{1}{2}\right)$ we can factor out,

Factor out 2 from the second parentheses,

So, by the Zero product property,

One root is $\frac{1}{2}$

The other two we get using the quadratic formula:

  $\begin{array}{l}{x}_{2,3}=\frac{1\pm \sqrt{1-4\cdot 2\cdot 6}}{2\cdot 2}=\frac{1\pm \sqrt{-47}}{4}\\ x_{2,3}=\frac{1\pm i\sqrt{-47}}{4}\end{array}$

Thus,

  $\begin{array}{l}{x}_1=\frac{1}{2}\\ x_{2,3}=\frac{1\pm i\sqrt{47}}{4}\end{array}$