Worked examples · Sheet 2

Equations of the form |A| = B

Theory used: Equation of the form |A| = B

Exercise 1

Solve the equation

∣ x - 2∣ = 3 x .

The right-hand side contains the variable, so the equation is equivalent to the union of two systems:

{3 x \geq 0 x - 2 = 3 x \cup {3 x \geq 0 x - 2 = - 3 x

First system.

{x \geq 0 - 2 x = 2 {x \geq 0 x = - 1

The system has no solutions.

Second system.

{x \geq 0 4 x = 2 ⎩ ⎨ ⎧ x \geq 0 x = \frac{1}{2}

The solution is $x = \frac{1}{2}$ .

The left-hand side is an absolute value and is never negative, so the right-hand side must also satisfy $3 x \geq 0$ : this condition enters both systems. In the first system the equation gives $x = - 1$ , which is negative, therefore the condition $x \geq 0$ is not satisfied and the value is discarded. In the second system the equation gives $x = \frac{1}{2}$ , which is positive, therefore the condition is satisfied. Checking by substitution confirms the result: $\frac{1}{2} - 2 = \frac{3}{2}$ and $3 \cdot \frac{1}{2} = \frac{3}{2}$ .

Exercise 2

Solve the equation

∣2 x + 1∣ = x + 4.

{x + 4 \geq 0 2 x + 1 = x + 4 \cup {x + 4 \geq 0 2 x + 1 = - (x + 4)

First system.

{x \geq - 4 x = 3

The solution $x = 3$ is acceptable.

Second system.

{x \geq - 4 3 x = - 5 ⎩ ⎨ ⎧ x \geq - 4 x = - \frac{5}{3}

The solution $x = - \frac{5}{3}$ is acceptable.

The solutions are $x = - \frac{5}{3}$ and $x = 3$ .

The acceptability condition is $x + 4 \geq 0$ , that is $x \geq - 4$ , and it applies to both systems. The first system gives $x = 3$ and the second gives $x = - \frac{5}{3}$ ; both values are greater than $- 4$ , so neither of them is discarded. Checking by substitution confirms the result: for $x = 3$ we have $∣7∣ = 7 = 3 + 4$ , and for $x = - \frac{5}{3}$ we have $- \frac{7}{3} = \frac{7}{3} = - \frac{5}{3} + 4$ .

Exercise 3

Solve the equation

∣ x^{2} - 1∣ = x + 1.

{x + 1 \geq 0 x^{2} - 1 = x + 1 \cup {x + 1 \geq 0 x^{2} - 1 = - (x + 1)

First system.

{x \geq - 1 x^{2} - x - 2 = 0 {x \geq - 1 (x - 2) (x + 1) = 0

The roots $x = 2$ and $x = - 1$ both satisfy the condition $x \geq - 1$ .

Second system.

{x \geq - 1 x^{2} + x = 0 {x \geq - 1 x (x + 1) = 0

The roots $x = 0$ and $x = - 1$ both satisfy the condition $x \geq - 1$ .

The solutions are $x = - 1$ , $x = 0$ and $x = 2$ .

The argument of the absolute value is quadratic, so each system contains a quadratic equation. In the first system the equation $x^{2} - x - 2 = 0$ factors as $(x - 2) (x + 1) = 0$ ; in the second the equation $x^{2} + x = 0$ is factored by taking out $x$ . All the roots found satisfy the condition $x \geq - 1$ , and the value $x = - 1$ appears in both systems, but in the union it is counted only once. Checking by substitution confirms the result: for $x = - 1$ both sides equal $0$ , for $x = 0$ they equal $1$ , for $x = 2$ they equal $3$ .