Solve

algebraically. Then compare the solution with that of Example 4. Noting that the algebraic steps after isolating the radical are identical, why is the solution different?

Data from Example 4

Solve the equation

We first isolate the radical by subtracting 3 from each side. This gives us

We now square both sides and proceed with the solution:

The solution x = 5 checks, but the solution x = 2 gives 4 = 2. Thus, the only solution is x = 5. The value x = 2 is an extraneous root. The graph of

shown in Fig. 14.19, verifies that x = 5 is a solution but x = 2 is not.

x 1 + x = 3