Worked examples · Sheet 3

Inequalities of the form √A < B

Theory used: Inequality of the form √A < B Domain conditions for roots of even index

Exercise 1

Solve the inequality

x - 1 < 2.

The inequality is equivalent to the system

⎩ ⎨ ⎧ x - 1 \geq 0 2 > 0 x - 1 < 4 {x \geq 1 x < 5

The solution is $1 \leq x < 5$ .

The solving system contains the domain condition $x - 1 \geq 0$ , the positivity of the right-hand side and the comparison between the radicand and the square of the right-hand side. The right-hand side is the constant $2$ , which is positive, so the condition $2 > 0$ is satisfied for every $x$ and excludes nothing. The conditions $x \geq 1$ and $x < 5$ remain, and they hold together on the interval indicated. Checking with a test value confirms the result: for $x = 2$ we have $1 = 1 < 2$ , while for $x = 10$ we have $9 = 3 > 2$ , and indeed $10$ does not belong to the solution.

Exercise 2

Solve the inequality

x + 6 \leq x .

⎩ ⎨ ⎧ x + 6 \geq 0 x \geq 0 x + 6 \leq x^{2} ⎩ ⎨ ⎧ x \geq - 6 x \geq 0 x^{2} - x - 6 \geq 0

The second-degree inequality factors:

(x - 3) (x + 2) \geq 0

x \leq - 2 \lor x \geq 3

The intersection with the conditions $x \geq - 6$ and $x \geq 0$ gives the solution $x \geq 3$ .

The inequality is non-strict, so the right-hand side must satisfy $x \geq 0$ and the comparison between radicand and square is non-strict as well. The inequality $x^{2} - x - 6 \geq 0$ has roots $- 2$ and $3$ and is satisfied outside the interval between the roots. Of the three conditions, $x \geq 0$ eliminates the half-line $x \leq - 2$ , therefore the half-line $x \geq 3$ remains, which also satisfies $x \geq - 6$ . Checking at the endpoint and with a test value confirms the result: for $x = 3$ we have $9 = 3 \leq 3$ , while for $x = 2$ we have $8 > 2$ , since $8 > 4$ , and indeed $2$ does not belong to the solution.

Exercise 3

Solve the inequality

2 x + 5 < x + 1.

⎩ ⎨ ⎧ 2 x + 5 \geq 0 x + 1 > 0 2 x + 5 < (x + 1)^{2} ⎩ ⎨ ⎧ x \geq - \frac{5}{2} x > - 1 2 x + 5 < x^{2} + 2 x + 1 ⎩ ⎨ ⎧ x \geq - \frac{5}{2} x > - 1 x^{2} > 4

The condition $x^{2} > 4$ is equivalent to $x < - 2 \lor x > 2$ ; the intersection with $x > - 1$ gives the solution $x > 2$ .

The right-hand side contains the variable, so the condition $x + 1 > 0$ enters the system alongside the domain condition. In the expansion of $(x + 1)^{2}$ the $2 x$ terms cancel with those on the left-hand side and the condition $x^{2} > 4$ remains, satisfied outside the closed interval with endpoints $- 2$ and $2$ . Since the condition $x > - 1$ excludes the half-line $x < - 2$ , the solution is the half-line $x > 2$ , which also satisfies the domain condition. Checking with a test value confirms the result: for $x = 3$ we have $11 < 4$ , since $11 < 16$ , while for $x = 2$ we have $9 = 3$ , which is not less than $3$ .