Page 152 - Math Course 2 (Book 1)

Radical Equations

Mo. 5

Lesson 3 x = 7 x = 2 Solve

x + 2 = x – 4 Original Equation
KEY CONCEPTS: ?
1. Solve radical equations. 7 + 2 = 7 – 4 x = 7
?
2. Solve radical equations with extraneous 9 = 3 Simplify
solutions.
3 = 3 True
?
x + 2 = x – 4 Original Equation
MO. 5 - L3a
?
2 + 2 = 2 – 4 x = 2
Solving Radical Equations 4 = – 2 Simplify.

Vocabulary A-Z

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Radical Equations Variable in Radical

h
Equations like t = that contain radicals with FREE-FALL HEIGHT
4
variables in the radicand are called radical equations. An object is dropped from an unknown
height and reaches the ground in 5 seconds.
3x + 12 = 3 3 x = 0 6 – x Use the equation to f nd the height from
h
t =
4
2c – 4 = 8 x = x + 20 which the object was dropped.
h
4b + 1 –3 = 0 5x – 6 = x Use the equation and replace t with
t =
5 seconds 4
Extraneous Solution h Original equation
t =
4
An extraneous solution is a solution derived from an
h
equation that is not a solution of the original 5 = Replace t with 5.
4
equation.

Squaring each side of an equation sometimes 20 = h Multiply each side by 4.
produces extraneous solutions.
2
(20) = ( h ) 2 Square each side.
2
( x + 2 ) = (x – 4) 2 Square each side.
2
x + 2 = x – 8x + 16 Simplify. 400 = h Simplify
2
0 = x – 9x + 14 Subtract x and 2 from Check
each side by substituting 400 for h in the original equation.

0 = (x – 7)(x – 2) Factor
Answer 400 ft
x – 7 = 0 or x – 2 = 0 Zero Product Property
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