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Worked solution

Four inequalities: zeros, poles and intersections

Without using a calculator, find the solution sets of the following inequalities.

(a) (x+2)2x2−1≤0\dfrac{(x+2)^2}{x^2-1}\leq0,

(b) x2+2x+2x2−x+2<0\dfrac{x^2+2x+2}{x^2-x+2}<0,

(c) 1(x−1)(x−3)<−1\dfrac1{(x-1)(x-3)}<-1,

(d) 1≤13x−2<31\leq\dfrac1{3x-2}<3.

CJC · Discussion · Question 1

Answer

(a) {−2}∪(−1,1)\{-2\}\cup(-1,1); (b) ∅\varnothing; (c) (1,2)∪(2,3)(1,2)\cup(2,3); (d) (7/9,1](7/9,1].

(a)

Factor the denominator: (x+2)2(x−1)(x+1)≤0\dfrac{(x+2)^2}{(x-1)(x+1)}\leq0. The critical values are −2,−1,1-2,-1,1. The sign is positive, positive, negative, positive on the four successive open intervals. At x=−2x=-2 the numerator is zero and the denominator is nonzero; include this isolated input. The denominator zeros ±1\pm1 are excluded.

The solution set is {x∈R:−1<x<1 or x=−2}\{x\in\mathbb R:-1<x<1\text{ or }x=-2\}.

How did this part go?

(b)

Complete the square in both polynomials:

(x+1)2+1(x−12)2+74<0\dfrac{(x+1)^2+1}{(x-\frac12)^2+\frac74}<0.

Both numerator and denominator are positive for every real input. The solution set is ∅\varnothing.

How did this part go?

(c)

Move the constant to the left:

1(x−1)(x−3)+1=x2−4x+4(x−1)(x−3)=(x−2)2(x−1)(x−3)<0\dfrac1{(x-1)(x-3)}+1=\dfrac{x^2-4x+4}{(x-1)(x-3)}=\dfrac{(x-2)^2}{(x-1)(x-3)}<0.

The denominator is negative between 1 and 3. At 2 the expression is zero, which fails the strict inequality. Thus {x∈R:1<x<3, x≠2}\{x\in\mathbb R:1<x<3,\ x\ne2\}.

How did this part go?

(d)

Solve both inequalities and intersect their sets.

13x−2≥1  ⟺  x−13x−2≤0\dfrac1{3x-2}\geq1\iff\dfrac{x-1}{3x-2}\leq0, giving 23<x≤1\frac23<x\leq1.

13x−2<3  ⟺  9x−73x−2>0\dfrac1{3x-2}<3\iff\dfrac{9x-7}{3x-2}>0, giving x<23x<\frac23 or x>79x>\frac79.

Their intersection is {x∈R:79<x≤1}\{x\in\mathbb R:\frac79<x\leq1\}.

How did this part go?

Compare your sketch

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Part (a) has signs positive, positive, negative, positive across −2, −1 and 1; −2 is included and ±1 are excluded. Part (c) has signs positive, negative, negative, positive across 1, 2 and 3; all three are excluded. In (d), intersect (2/3,1] with the two rays x<2/3 or x>7/9 to obtain (7/9,1].
Part (a) has signs positive, positive, negative, positive across −2, −1 and 1; −2 is included and ±1 are excluded. Part (c) has signs positive, negative, negative, positive across 1, 2 and 3; all three are excluded. In (d), intersect (2/3,1] with the two rays x<2/3 or x>7/9 to obtain (7/9,1].
Part (a) has signs positive, positive, negative, positive across −2, −1 and 1; −2 is included and ±1 are excluded. Part (c) has signs positive, negative, negative, positive across 1, 2 and 3; all three are excluded. In (d), intersect (2/3,1] with the two rays x<2/3 or x>7/9 to obtain (7/9,1].

Common mistakes

A repeated numerator root need not change the sign, but it can still be an included isolated solution.

Solution notes

The solution diagrams are independently drawn from the printed mathematical relationships. Geometry is illustrative; use the labelled critical values and endpoint symbols.

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