Porisma

Worked solution

A rational identity and a scaled input

Show that 2x2+2x−8x2+x−6−1=x2+x−2x2+x−6\dfrac{2x^2+2x-8}{x^2+x-6}-1=\dfrac{x^2+x-2}{x^2+x-6}. mark

Hence, without using a calculator, solve the inequality 2x2+2x−8x2+x−6≤1\dfrac{2x^2+2x-8}{x^2+x-6}\leq1. marks

Deduce the solution to the inequality 4x2+2x−42x2+x−3≤1\dfrac{4x^2+2x-4}{2x^2+x-3}\leq1. marks

[2016/PJC/Promo/3]

CJC · Review Problems · Question 2

Answer

Base: (−3,−2]∪[1,2)(-3,-2]\cup[1,2). Deduction: (−3/2,−1]∪[1/2,1)(-3/2,-1]\cup[1/2,1).

Show

Subtract over the common denominator: (2x2+2x−8)−(x2+x−6)x2+x−6=x2+x−2x2+x−6\dfrac{(2x^2+2x-8)-(x^2+x-6)}{x^2+x-6}=\dfrac{x^2+x-2}{x^2+x-6}, for x≠−3,2x\ne-3,2.

How did this part go?

Hence

The inequality becomes (x−1)(x+2)(x+3)(x−2)≤0\dfrac{(x-1)(x+2)}{(x+3)(x-2)}\leq0. The successive signs across −3,−2,1,2-3,-2,1,2 are +,−,+,−,++,-,+,-,+. Include the numerator zeros −2,1-2,1 and exclude both poles. Hence −3<x≤−2-3<x\leq-2 or 1≤x<21\leq x<2.

How did this part go?

Deduce

Multiply numerator and denominator of the target fraction by 2 to obtain 8x2+4x−84x2+2x−6\dfrac{8x^2+4x-8}{4x^2+2x-6}. This is the original fraction with input 2x2x. Thus −3<2x≤−2-3<2x\leq-2 or 1≤2x<21\leq2x<2, giving −32<x≤−1-\frac32<x\leq-1 or 12≤x<1\frac12\leq x<1.

How did this part go?

Compare your sketch

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The factored difference has signs positive, negative, positive, negative, positive across −3, −2, 1, 2. Poles −3 and 2 are open; zeros −2 and 1 are included.
The factored difference has signs positive, negative, positive, negative, positive across −3, −2, 1, 2. Poles −3 and 2 are open; zeros −2 and 1 are included.
The factored difference has signs positive, negative, positive, negative, positive across −3, −2, 1, 2. Poles −3 and 2 are open; zeros −2 and 1 are included.

Common mistakes

An identity involving fractions holds only where their denominator is nonzero.

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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