KindiakMath

Solving Inequalities

Question 1. Without the use of graphs, solve the inequality

(x-1)(x-3) \geq 0.

(Click for Solution)

Solution. The inequality breaks down into two cases:

(x-1)(x-3) = 0 \quad \text{or} \quad (x-1)(x-3) > 0.

For the first case, since the right-hand side is zero, at least one factor must be zero. Therefore,

x-1 = 0 \ \text{or}\ x-3 = 0\quad \Rightarrow \quad x = 1 \ \text{or} \ 3.

For the second case, we want to solve

(x-1)(x-3) > 0.

If x = 1, then the left-hand side reduces to 0, a contradiction. Therefore, only two possible cases remain: x < 1 or x > 1. Suppose firstly that x > 1, so that x-1 > 0. Dividing both sides by x-1 preserves the inequality direction:

\displaystyle x - 3 = \frac{(x-1)(x-3)}{x-1} > \frac{0}{x-1} = 0,

so that x > 3. Taking the stricter condition between x > 1 and x > 3, we obtain x > 3.

Suppose secondly that x < 1, so that x - 1 < 0. Dividing both sides by x-1 reverses the inequality direction:

\displaystyle x - 3 = \frac{(x-1)(x-3)}{x-1} < \frac{0}{x-1} = 0,

so that x < 3. Taking the stricter condition between x < 1 and x < 3, we obtain x < 1.

Therefore, the four possibilities are

x = 1, \quad x < 1,\quad x = 3,\quad x > 3.

Collapsing the inequalities yields our final answer: x \leq 1 or x \geq 3.

Question 2. With the use of graphs, solve the inequality

(x-1)(x-3) \geq 0.

(Click for Solution)

Solution. Expanding the left-hand side and completing the square,

\begin{aligned} (x-1)(x-3) &= x^2 - 3x - x + 3 \\ &= x^2 - 4x + 3 \\ &= (x^2 - 2 \cdot 2 \cdot x + 2^2) - 2^2 + 3 \\ &= (x-2)^2 - 1,\end{aligned}

we obtain a upwards-directed quadratic curve with x-intercepts 1, 3, as computed in Question 1.

The inequality requires that y \geq 0. Hence, the desired x-values are

x \leq 1 \quad \text{or} \quad x \geq 3.

Remark 1. It is more often easier and much faster to solve using graphs than without.

Question 3. Given x \neq 0, solve the inequality

4x \leq x^2 + 3.

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Solution. Doing algebra and factorising,

\begin{aligned}x^2 + 3 &\geq 4x \\ x^2 - 4x + 3 &\geq 0 \\ (x-1)(x-3) & \geq 0. \end{aligned}

By Question 1 or 2, x \leq 1 or x \geq 3.

—Joel Kindiak, 17 Jun 26, 1614H

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joelkindiak

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