Question

Suppose that the equation $a{x}^2+bx+c=0$has real coefficients and complex roots. Why must the roots be complex conjugates of each other ? [Hint: Think about how you would find the roots using the Quadratic Formula.]

Short answer

The complex roots of a quadratic equation are complex conjugate of each other.

Step-by-step solution

Step 1. Concept of complex conjugate.

The complex conjugate of the complex number$a+bi$ is represented by a horizontal bar over it, that is, the complex conjugate of$a+bi$ is denoted by $\overline{a+bi}$.

The complex conjugate of any complex number can be obtained by changing the sign of the imaginary part of the complex number. The complex conjugate of $a+bi$is $a-bi$.

The product of a complex number and its complex conjugate is given by

$\left(a+bi\right)\left(a-bi\right)=a^2+b^2$

The imaginary number i is defined as the square root of negative one. That is, $i^2=-1$.

Step 2. Concept of imaginary number.

The imaginary number i is defined as the square root of negative one. That is, $i=\sqrt{-1}$or $i^2=-1$.

The solutions of the quadratic equation $a{x}^2+bx+c=0$, where a, b, and c are constants, is given by the quadratic formula

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

Step 3. Solve the quadratic equation.

The roots of the quadratic equation $a{x}^2+bx+c=0$, where a, b, and c are real constants, is given by the quadratic formula

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

The roots of this equation will be complex only when the discriminant$b^2-4ac$ is negative.

Two roots of this equation obtained by taking the sign in front of radical sign as positive and negative with each of two roots.

So, if the roots are imaginary then one root will have the imaginary part with positive sign and another root must be with negative sign.

So, by the definition of complex conjugates, the complex solutions of the quadratic equation will be complex conjugates of each other.