Porisma

Worked solution

A positive quadratic with an absolute-value input

Without using a calculator, solve the inequality 3x2x2−3x+1>−1\dfrac{3x}{2x^2-3x+1}>-1. marks

Hence solve 3∣x∣2x2−3∣x∣+1>−1\dfrac{3|x|}{2x^2-3|x|+1}>-1. marks

[2016/YJC/Promo/10(b)]

CJC · Review Problems · Question 7

Answer

Base: x<12x<\frac12 or x>1x>1. Hence (−∞,−1)∪(−1/2,1/2)∪(1,∞)(-\infty,-1)\cup(-1/2,1/2)\cup(1,\infty).

First inequality

Add 1 to obtain 2x2+12x2−3x+1>0\dfrac{2x^2+1}{2x^2-3x+1}>0. The numerator is positive for all real xx, so solve 1(x−1)(2x−1)>0\dfrac1{(x-1)(2x-1)}>0. The signs across 12,1\frac12,1 are +,−,++,-,+, with both poles excluded. Hence x<12x<\frac12 or x>1x>1.

How did this part go?

Hence

Since x2=∣x∣2x^2=|x|^2, replace the original input by ∣x∣|x|. Then ∣x∣<12|x|<\frac12 or ∣x∣>1|x|>1, giving −12<x<12-\frac12<x<\frac12 or x<−1x<-1 or x>1x>1.

How did this part go?

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The reciprocal of (x−1)(2x−1) is positive to the left of 1/2 and right of 1, and negative between them. Both poles are excluded.
The reciprocal of (x−1)(2x−1) is positive to the left of 1/2 and right of 1, and negative between them. Both poles are excluded.
The reciprocal of (x−1)(2x−1) is positive to the left of 1/2 and right of 1, and negative between them. Both poles are excluded.

Common mistakes

The x² term remains correct under the replacement because x²=|x|².

Solution notes

The sign chart is independently drawn from the printed solution. Spacing is illustrative; labelled critical values and open/closed endpoints determine the set.

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