# When I complete the square on $3x^2 - 12x + 14$ I get an imaginary number, where have I gone wrong?

I have a question in my excersise book:

By completing the square show that the expression $$3x^2 - 12x + 14$$ is positive for all $$x$$

My approach was to complete the square and rearrange to make $$x$$ the subject.

The answer I came to after completing the square was $$(\sqrt {3}x - 2\sqrt{3})^2+2$$.

However I get a negative square root:

$$(\sqrt {3}x - 2\sqrt{3})^2+2 = 0$$ $$(\sqrt {3}x - 2\sqrt{3})^2 = -2$$ $$\sqrt {3}x - 2\sqrt{3} = \sqrt{-2}$$ $$\sqrt{3}x = 2\sqrt{3} +- \sqrt {-2}$$ $$x = (2\sqrt{3} +- \sqrt {-2})/3$$

Bad formatting: $$+-$$ means either $$+$$ or $$-$$

Where have I gone wrong?

You haven't gone wrong per se. You've just gone a step too far. No need to solve the equation or factor anything. Just note that when you have $$(\sqrt 3x - 2\sqrt3)^2 + 2$$ then that's a square (which is non-negative) plus $$2$$, which necessarily makes the value of the entire expression strictly positive, no matter what $$x$$ is.
You haven't gone wrong. By finding the expression for $$x$$ after completing the square, you are looking for solutions to the equation $$3x^2 - 12x + 14 = 0$$. You find that there are no (real) solutions, which means that the graph of this parabola never touches the $$x$$-axis. Because this is a "valley parabola", certainly it will be positive somewhere; that means it is always positive, because it never crosses or touches the $$x$$-axis.
Do you have to do it exactly that way? Another is to take the factor of 3 (the coefficient of $$x^2$$) outside, & put it back at the end, to get $$3(x-2)^2+2 .$$ Certainly, this shows just as well that the expression is always positive.
In general, remember that a negative radicand does not imply you went wrong anywhere. If anything, it is simply another way to point out that the graph never crosses the $$x$$-axis (hence no real roots), and since $$a > 0$$, the graph lies entirely above the $$x$$-axis, which means the quadratic is always positive.