Ask Uncle Colin: An absolute quadratic inequality

Dear Uncle Colin,

I have to solve the inequality $x^{2} - | 5 x - 3 | < 2 + x$ . I rearranged to make it $x^{2} - x - 2 < | 5 x - 3 |$ , but the final answer is eluding me.

-- Put Right Inequality Muddle

Hello, PRIM!

You’re off to a good start; the next thing I would do would be to sketch the two curves. Here’s Desmos’s version:

We want to know where the left-hand side (the curve) is below the right-hand side (the V). That’s clearly between the two crossing points; now all you need to do is find the crossing points!

On the right, the modulus graph is $y = 5 x - 3$ (it’s sloping upwards, so it must be the version with the positive gradient). You need to solve $x^{2} - x - 2 = 5 x - 3$ , or $x^{2} - 6 x + 1 = 0$ . That has roots at $3 \pm 2 \sqrt{2}$ , but the value you want is greater than $x = \frac{3}{5}$ because that’s where the point of the V is. You need the positive solution, which is $x = 3 + 2 \sqrt{2}$ .

On the left, the graph is $y = 3 - 5 x$ and you need to solve $x^{2} - x - 2 = 3 - 5 x$ , or $x^{2} + 4 x - 5 = 0$ . That factorises as $(x + 5) (x - 1) = 0$ , and so $x = - 5$ or $x = 1$ . However, $x = 1$ doesn’t fit with the graph (for this one, $x < \frac{3}{5}$ ), so the solution you want is $x = - 5$ .

That means the final answer you need is $- 5 < x < 3 + 2 \sqrt{2}$ .

Hope that helps!

-- Uncle Colin

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