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Worked solution

Prove a quadratic is positive before removing it

For real x, prove 2x2−x+22x^2-x+2 is always positive. Solve (3x−1)/(2x2−x+2)≥1/x(3x-1)/(2x^2-x+2)\ge1/x, leaving the answer in exact form.

[2013/NJC/I/2(part)]

CJC · Lecture · Example 5

Positivity method 1: complete the square

2x2−x+2=2(x−1/4)2+15/8≥15/8>02x^2-x+2=2(x-1/4)^2+15/8\ge15/8>0, proving positivity for every real x.

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Positivity method 2: use the discriminant

The discriminant is b2−4ac=1−4(2)(2)=−15<0b^2-4ac=1-4(2)(2)=-15<0, and the coefficient of x² is positive. The upward-opening quadratic never meets the axis, so it is always positive.

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Combine the fractions

Move the right side to the left: (3x−1)/(2x2−x+2)−1/x≥0(3x-1)/(2x^2-x+2)-1/x\ge0. The common numerator is 3x2−x−2x2+x−2=x2−23x^2-x-2x^2+x-2=x^2-2, so require (x2−2)/[x(2x2−x+2)]≥0(x^2-2)/[x(2x^2-x+2)]\ge0, with x≠0.

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Remove only the positive factor and read signs

Since the quadratic is strictly positive, this is equivalent to (x2−2)/x≥0(x^2-2)/x\ge0, or (x−2)(x+2)/x≥0(x-\sqrt2)(x+\sqrt2)/x\ge0. Mark −√2,0,√2; interval signs are −,+,−,+. Hence −2≤x<0-\sqrt2\le x<0 or x≥2x\ge\sqrt2. Keep the numerator zeros and exclude the pole 0.

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Compare your sketch

Expand diagram

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The Teacher-completed sign line marks−√2 and√2 as included numerator zeros and 0 as an excluded denominator zero. Signs are minus, plus, minus, plus.
The Teacher-completed sign line marks−√2 and√2 as included numerator zeros and 0 as an excluded denominator zero. Signs are minus, plus, minus, plus.
The Teacher-completed sign line marks−√2 and√2 as included numerator zeros and 0 as an excluded denominator zero. Signs are minus, plus, minus, plus.

Common mistakes

It is safe to remove 2 x²−x+2 only after proving its sign. The remaining factor x still changes sign.

Read the hint