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Inequalities with squares

The inequality $x^2 \ge k$ with $k > 0$ has two parts, because both a positive and a negative number can have the same square. You find the two zeros $\sqrt{k}$ and $-\sqrt{k}$ and take the region outside them. Many people forget the negative part and answer only $x \ge \sqrt{k}$ . If $k = 0$ , the solution is all $x$ . If $k < 0$ , the inequality holds for all $x$ .

x^2 \ge k \iff x \le -\sqrt{k}\ \text{or}\ x \ge \sqrt{k}

when

k > 0

x^2 + 1 > 0

true for every real x

Symbols

k

a positive number

Example

$x^2 \ge 9$ . The zeros are $x = 3$ and $x = -3$ , since $3^2 = 9$ and $(-3)^2 = 9$ . The solution is $x \le -3$ or $x \ge 3$ .

Never write

x \ge \pm 3

. That is not a valid interval.

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Part of Mathematics 1T: Inequalities and sign charts.