Porisma

Worked solution

Recognise the same rational inequality after a substitution

Without using a calculator, solve (x2+2x+8)/[(x+3)(x−2)]<0(x^2+2x+8)/[(x+3)(x-2)]<0.

Hence solve (x2+2∣x∣+8)/[(∣x∣+3)(∣x∣−2)]<0(x^2+2|x|+8)/[(|x|+3)(|x|-2)]<0.

CJC · Lecture · Example 10

Complete the square and remove a known positive numerator

x2+2x+8=(x+1)2+7>0x^2+2x+8=(x+1)^2+7>0 for every real x. Thus the first inequality is equivalent to 1/[(x+3)(x−2)]<01/[(x+3)(x-2)]<0. Mark poles −3,2; signs are +,−,+. Hence −3<x<2.

How did this part go?

Match every occurrence in the second expression

The repeated blank in the source prompt is filled by |x|: its square is |x|²=x², its linear term is 2|x| and both denominator factors use|x|. Replace the old variable by u=|x| to obtain −3<|x|<2.

How did this part go?

Apply the range of the substituted expression

Since |x|≥0, the lower bound −3<|x| always holds. The graph of y=|x| with levels −3 and 2 leaves |x|<2, so −2<x<2. The inputs ±2 are poles in the substituted expression and are excluded.

How did this part go?

Compare your sketch

Expand diagram

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The original sign diagram has plus, minus, plus with excluded poles −3 and 2. The modulus graph compares|x|with both source levels 2 and −3 and retains crossing inputs −2 and 2.
The original sign diagram has plus, minus, plus with excluded poles −3 and 2. The modulus graph compares|x|with both source levels 2 and −3 and retains crossing inputs −2 and 2.
The original sign diagram has plus, minus, plus with excluded poles −3 and 2. The modulus graph compares|x|with both source levels 2 and −3 and retains crossing inputs −2 and 2.

Common mistakes

A valid hence substitution must replace every occurrence, including squared terms. Use |x|²=x² and then apply |x|≥0.

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