# Ask Uncle Colin: A quadratic inequality

Dear Uncle Colin,

I’ve come across a seemingly simple question I can’t tackle: solve $x^2 + 2x \ge 2$.

I tried factorising to get $x(x+2) \ge 2$, which has the roots 0 and -2, but the book says the answer is $x < -1-\sqrt{3}$ or $x > -1 + \sqrt{3}$. Where have I gone wrong?

-- Running Out Of Time

Hi, ROOT!

First thing to do, when you have a disagreement over an answer, is to check that yours makes sense. If $x=0$, does the inequality work? No, you have $0 \ge 2$ for both of your answers, which is clearly wrong.

Also, as a rule (except for a few contrived edge cases), if you’re solving an inequality, you would normally expect a range of answers rather than specific values – and this example isn’t one of the exceptions!

What you’ve done is solve $x^2 + 2x = 0$, which is a different problem altogether.

If I were doing this ((and what do you know? I *am* doing this!)) , I would bring everything to the left:

$x^2 + 2x - 2 \ge 0$, then complete the square:

$(x+1)^2 - 3 \ge 0$

At this point, I’d sketch the curve $y = (x+1)^2 - 3$. It has a vertex at $(-1, -3)$ and crosses the $x$-axis at … well, let’s work out where $(x+1)^2 - 3 = 0$.

Adding three to each side gives $(x+1)^2 = 3$, so $x+1 = \pm \sqrt{3}$, giving $x = -1\pm \sqrt{3}$ as the crossing points.

To find where the expression is at least 0, we need to find the $x$-values associated with non-negative $y$-values – we need to be the left of the lower root ($x \le -1-\sqrt{3}$) or to the right of the upper root ($x \ge -1 + \sqrt{3}$).

Hope that helps!

-- Uncle Colin