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

cubicsquadraticssquare-numbers

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?

Best Answer

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.

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