Chapter 8.6, Problem 24WE

a.

To determine

To show that the given number is a root of the equation.

Explanation of Solution

Given information:

The given equation is $x^3-2x^2+2x+5i=0$ , and the given number is $2-i$

Concept used:

If a number is the root of a certain equation then it must satisfy the equation.

Proof:

Substitute the given number in the left hand side of equation

  $\begin{array}{l}{x}^3-2x^2+2x+5i\\ {\left(2-i\right)}^3+2{\left(2-i\right)}^2+2\left(2-i\right)+5i\\ \left(2-11i\right)-2\left(3-4i\right)+4-2i+5i\\ 2-11i-6+8i+4-2i+5i\\ 0\end{array}$

The number satisfies the equation hence it is a root of the equation.

b.

To determine

To show that the conjugate of given number is not a root of the equation.

Explanation of Solution

Given information:

The given equation is $x^3-2x^2+2x+5i=0$ , and the given number is $2-i$

Proof:

Conjugate of $2-i$ is $2+i$

Substitute the conjugate in the left hand side of equation

  $\begin{array}{l}{x}^3-2x^2+2x+5i\\ {\left(2+i\right)}^3-2{\left(2+i\right)}^2+\left(2+i\right)+5i\\ \left(2+11i\right)-2\left(3+4i\right)+2\left(2+i\right)+5i\\ 2+11i-6-8i+4+2i+5i\\ 10i\end{array}$

10i is not equals to zero hence is not satisfies the equation.

The conjugate of the number does not satisfies the equation hence it is not a root of the equation.