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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?

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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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  • $\begingroup$ Thank you. That was much easier than I expected it to be, I'll accept your answer as soon as I can $\endgroup$ – Simon Dec 3 '18 at 12:21
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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.

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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.

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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.

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