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

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Solution

Finding the root of the equation:
We know for the quadratic equation $a x^{2} + b x + c = 0$ if
$b^{2} - 4 a c > 0$ then there are two roots the roots are real
$b^{2} - 4 a c = 0$ then roots are real and of order two
$b^{2} - 4 a c < 0$ then there are two roots the roots are complex root
Given quadratic equation is $3 x^{2} + 3 x + 3 = 0$
Now comparing with the standard form we have $a = 3, b = 3, c = 3 .$
$∴ b^{2} - 4 a c = 3^{2} - 4 \cdot 3 \cdot 3 = 9 - 36 = - 27 ∴ b^{2} - 4 a c < 0$
Since the value of $(b^{2} - 4 a c)$ is negative, $3 x^{2} + 3 x + 3 = 0$ has $2$ complex roots which are complex conjugates of each other.
Hence the polynomial $3 x^{2} + 3 x + 3 = 0$ has $2$ complex roots.

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