Worked examples · Sheet 1

Equations of the form √A = B

Theory used: Equation of the form √A = B Domain conditions for roots of even index

Exercise 1

Solve the equation

2 x - 1 = 3.

The equation is equivalent to the system

{3 \geq 0 2 x - 1 = 9 {3 \geq 0 x = 5

The solution is $x = 5$ .

The right-hand side is the constant $3$ , which is nonnegative, so the condition $B \geq 0$ is satisfied for every $x$ and only the equation between the radicand and the square of the right-hand side remains. From $2 x - 1 = 9$ it follows that $x = 5$ . Checking by substitution confirms the result, since the radicand equals $2 \cdot 5 - 1 = 9$ and $9 = 3$ .

Exercise 2

Solve the equation

x + 3 = x - 3.

{x - 3 \geq 0 x + 3 = (x - 3)^{2} {x \geq 3 x + 3 = x^{2} - 6 x + 9 {x \geq 3 x^{2} - 7 x + 6 = 0

The second-degree equation factors:

(x - 1) (x - 6) = 0

x = 1 \lor x = 6

The value $x = 1$ does not satisfy the condition $x \geq 3$ and is discarded. The solution is $x = 6$ .

The right-hand side contains the variable, so the condition $x - 3 \geq 0$ enters the system and decides which roots are acceptable. Squaring gives the equation $x^{2} - 7 x + 6 = 0$ , with roots $1$ and $6$ . Since $1 < 3$ , the first root does not satisfy the condition and is discarded; indeed, for $x = 1$ the left-hand side equals $4 = 2$ and the right-hand side equals $- 2$ . Checking by substitution confirms the remaining solution: for $x = 6$ we have $9 = 3 = 6 - 3$ .

Exercise 3

Solve the equation

x^{2} + 3 x = 2 x .

{2 x \geq 0 x^{2} + 3 x = 4 x^{2} {x \geq 0 3 x^{2} - 3 x = 0 {x \geq 0 3 x (x - 1) = 0

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

The solutions are $x = 0$ and $x = 1$ .

The acceptability condition is $2 x \geq 0$ , that is, $x \geq 0$ . Squaring, the radicand $x^{2} + 3 x$ must be set equal to $(2 x)^{2} = 4 x^{2}$ ; moving everything to one side gives $3 x^{2} - 3 x = 0$ , which factors by factoring out $3 x$ . Both roots are nonnegative, so neither is discarded. Checking by substitution confirms the result: for $x = 0$ both sides equal $0$ , and for $x = 1$ we have $1 + 3 = 2 = 2 \cdot 1$ .