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Worked solution

Prove positivity and retain an isolated equality

Given that xx is real, prove algebraically that x2−2x+3x^2-2x+3 is always positive. mark

Hence, without using a graphing calculator, find the range of values of xx that satisfy x+4x2−2x+3≥32−x\dfrac{x+4}{x^2-2x+3}\geq\dfrac3{2-x}. marks

[2016/MJC/Promo/1]

CJC · Review Problems · Question 6

Answer

x=12x=\frac12 or x>2x>2.

Positivity

Complete the square: x2−2x+3=(x−1)2+2>0x^2-2x+3=(x-1)^2+2>0 for every real xx. Alternatively, the discriminant is (−2)2−4(1)(3)=−8<0(-2)^2-4(1)(3)=-8<0 and the leading coefficient is positive, so the quadratic is positive everywhere.

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Inequality

The difference is (x+4)(2−x)−3(x2−2x+3)(x2−2x+3)(2−x)=−4x2+4x−1(x2−2x+3)(2−x)=−(2x−1)2(x2−2x+3)(2−x)≥0\dfrac{(x+4)(2-x)-3(x^2-2x+3)}{(x^2-2x+3)(2-x)}=\dfrac{-4x^2+4x-1}{(x^2-2x+3)(2-x)}=\dfrac{-(2x-1)^2}{(x^2-2x+3)(2-x)}\geq0. Divide out the strictly positive quadratic and negate to get (2x−1)22−x≤0\dfrac{(2x-1)^2}{2-x}\leq0. This is positive on both sides of 12\frac12 before the pole 2, zero at 12\frac12, and negative for x>2x>2. Thus x=12x=\frac12 or x>2x>2.

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Compare your sketch

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For (2x−1)²/(2−x), signs are positive, positive, negative across 1/2 and 2. The zero 1/2 is included as an isolated point; the pole 2 is excluded.
For (2x−1)²/(2−x), signs are positive, positive, negative across 1/2 and 2. The zero 1/2 is included as an isolated point; the pole 2 is excluded.
For (2x−1)²/(2−x), signs are positive, positive, negative across 1/2 and 2. The zero 1/2 is included as an isolated point; the pole 2 is excluded.

Common mistakes

Dividing by the squared numerator would lose the isolated solution x=1/2.

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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