Page 731 - Algebra 1

Math Language
The derived equation
is a new equation that results from squaring the original equation.
When both sides of an equation are squared to solve an equation, the resulting equation may have solutions that do not satisfy the original equation. Recall that an extraneous solution is a solution of a derived equation that does not satisfy the original equation.
Determining Extraneous Solutions
Solve each equation.
a. √ x- 1=x-3
SOLUTION The radical expression is isolated. Use inverse operations.
Example
4
√ x - 1 = x - 3 (√ x - 1 )2 = (x - 3)2
x-1=x2 -6x+9 +__1 = +__1
x=x2 -6x+10 -x = -x
0=x2 -7x+10
0 = (x - 2)(x - 5) x-2=0 x-5=0 x = 2 x = 5
Check √ x- 1 =x-3;x=2 √ 2 - 1 2 - 3
√ 1 - 1
1 ≠ -1 ✗
Square both sides.
Simplify.
Addition Property of Equality
Subtraction Property of Equality Simplify.
Factor.
Writetwoequations.
Use inverse operations to simplify.
√ x- 1 =x-3;x=5 √ 5 - 1 5 - 3
√ 4 2
2 = 2 ✓
Hint
Remember that positive numbers have two square roots. By convention, √ returns the positive square root.
The solution x = 2 is extraneous, so x = 5 is the only solution.
b . √ x + 5 = - 2
SOLUTION Use inverse operations to isolate the radical.
√ x + 5 = - 2 -__5 = -__5 √x = -7
(√x)2 = (-7)2
x = 49
Check √x + 5 = -2 √ 49 + 5 -2 7 + 5 -2 12 ≠ -2
Subtraction Property of Equality Simplify.
Square both sides.
Simplify.
✗
The solution x = 49 is extraneous. There is no solution.
716 Saxon Algebra 1