Ask Uncle Colin: An Absolute Quadratic Inequality

Dear Uncle Colin,

I'm pretty good with quadratic inequalities and pretty good with absolute values, but when I get the two together, I get confused. For example, I struggled with the set of values satisfying $x^2 -\left| 5x-3\right| < 2 + x$. Can you help?

– Nasty Absolute Value Inequalities Ending Rongly

Hi, NAVIER, and thanks for your message!

These are nasty, but can be made simpler by rearranging and sketching.

I would begin by getting the absolute value on one side and everything else on the other: $ x^2 – x – 2 < \left| 5x-3\right|$.

The left-hand side is a quadratic that cuts the $x$-axis at $(-1,0)$ and $(2,0)$; the right-hand side is a steep V-shape that bounces off of the $x$-axis at $x= \frac{3}{5}$. The quadratic curve is below the V between the two points where they cross.

I would now treat this as two inequalities. When $x < \frac{3}{5}$, the absolute value part is $3-5x$, and the two cross when $x^2 – x – 2 = 3 – 5x$. Rearranging, this gives $x^2 + 4x – 5=0$, which has solutions at $x=-5$ and $x=1$. Only the first of those is in the domain we're looking at, so one of the crossing points is where $x=-5$.

The other comes from solving $x^2 – x – 2 = 5x – 3$ for $x > \frac{3}{5}$, which simplifies to $x^2 – 6x + 1$. This has solutions at $x = 3 \pm 2\sqrt{2}$, and only the larger of those is in the domain (since $3 – 2\sqrt{2} \approx 0.2 < \frac{3}{5}$).

As we know the set of values that works is all of the $x$s between the crossing points, our final answer is $-5 \lt x \lt 3 + 2\sqrt{2}$.

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.


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