Chapter 3.5, Problem 81E

To determine

To explain: the quadratic equation with real coefficients and complex roots must have complex conjugate roots.

Explanation of Solution

A quadratic equation $a{x}^2+bx+c=0$ has two distinct solutions when the discriminant $D=0$ . The solutions of the quadratic equation $a{x}^2+bx+c=0$ is obtained using the formula,

  $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

This is equal to two distinct values for $x$ , by splitting the denominator for each numerator.

  $x=\frac{-b}{2a}\pm \frac{\sqrt{b^2-4ac}}{2a}$

By substituting the values of the variables $a,b$ and $c$ , get $x=z\pm yi$ , where $z$ and $y$ are my real value. That is the two solutions of $x$ are $x=z+yi$ and $x=z-yi$ .

Now, consider the roots one by one and find its conjugate.

First take $x=z+yi$ .

The complex conjugate of $x$ is $\overline{x}$ .

If $x=a+bi$ then the complex conjugate is $x=a-bi$

Therefore, the conjugate of $x=z+yi$ is $x=z-yi$ .

Hence, the roots are complex conjugate to each other.

Similarly, the complex conjugate of the other root $x=z-yi$ is $x=z+yi$ .