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 ¹ , 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

Footnotes:

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