Question

Find the zeros of the function.f(x) = 3x^2 + 6

Answer 1

f(x) = 3x^2 + 6 ​

0 = 3x^2 + 6 (Replacing f(x) by 0)

-6 = 3x^2 (Subtracting 6 from both sides of the equation)

-6/3 = x^2 (Dividing by 3 on both sides of the equation)

-2 = x^2 (Dividing)

`\begin{gathered} \sqrt[]{-2}\text{ = x (Taking the square root on both sides of the equation)} \\ \sqrt[]{-1}\text{ }\cdot\sqrt[]{2}\text{= x (Separating terms)} \\ \sqrt[]{2}i=x\text{ (Changing }\sqrt[]{-1}\text{ for i)} \end{gathered}`

There are no real solutions. The complex solutions are:

`\pm\sqrt[]{2}i`