Ask Uncle Colin: A quadratic inequality

Ask Uncle Colin is a chance to ask your burning, possibly embarrassing, maths questions -- and to show off your skills at coming up with clever acronyms. Send your questions to and Uncle Colin will do what he can.

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


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 this1 , 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


Colin is a Weymouth maths tutor, author of several Maths For Dummies books and A-level maths guides. He started Flying Colours Maths in 2008. He lives with an espresso pot and nothing to prove.

  1. and what do you know? I am doing this! []


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