Chapter 1.8, Problem 66E

Solution

To show: There is no quadratic equation $a{x}^2+bx+c=0$ such that $a,b,$ and $c$ are real numbers and $3i$ and $-2i$ are solutions.

Given information:

  $3i$ and $-2i$ are solutions.

Concept Used:

The sum of the roots of the quadratic equation $a{x}^2+bx+c=0$ where $a,b$ and $c$ are real constants must be equal to $-\frac{b}{a}$ .

Calculation:

Consider the quadratic equation $a{x}^2+bx+c=0$ where $a,b$ and $c$ are real constants.

The sum of the roots of $3i,-2i$ is $3i+\left(-2i\right)=i$ .

Since sum of the roots is $-\frac{b}{a}$

  $\begin{array}{c}i=-\frac{b}{a}\\ \frac{b}{a}=-i\end{array}$

Given b and a both are real numbers then $\frac{b}{a}$ must be a real number. But, $i$ is not a real number.

Hence, there is no quadratic equation $a{x}^2+bx+c=0$ that are real numbers and $3i,-2i$ are solutions.