Question

How many roots does the polynomial $3x^2+3x+3=0$ have?

Answer 1

Finding the root of the equation:

We know for the quadratic equation $a{x}^2+bx+c=0$ if

1. $b^2-4ac>0$ then there are two roots the roots are real
2. $b^2-4ac=0$ then roots are real and of order two
3. $b^2-4ac<0$ then there are two roots the roots are complex root

Given quadratic equation is $3x^2+3x+3=0$

Now comparing with the standard form we have $a=3,b=3,c=3.$

$$\begin{gathered} \therefore b^2-4ac \\ =3^2-4·3·3 \\ =9-36 \\ =-27 \\ \therefore b^2-4ac<0 \end{gathered}$$

Since the value of $(b^2-4ac)$ is negative, $3x^2+3x+3=0$ has $2$ complex roots which are complex conjugates of each other.

Hence the polynomial $3x^2+3x+3=0$ has $2$ complex roots.