3-4 Radical Equations (Presentation)

8/8/2019 3-4 Radical Equations (Presentation)

http://slidepdf.com/reader/full/3-4-radical-equations-presentation 1/14

3-4 Radical Equations

Unit 3 Quadratic and Polynomial Functions



Concepts and Objectives

Objective #13

Solve equations with radicals and check the solutions Solve equations that are quadratic in form



Power Property

If P and Q are algebraic expressions, then every

solution of the equation P = Q is also a solution of the equation P n = Qn, for any positive integer n.

Note: This does not mean that every solution of P n = Qn

is a solution of P = Q.

We use the power property to transform an equation

that is difficult to solve into one that can be solved more

easily. Whenever we change an equation, however, it is

essential to check all possible solutions in the original

equation.



Solving Radical Equations

Step 1 Isolate the radical on one side of the equation.

Step 2 Raise each side of the equation to a power that isthe same as the index of the radical to eliminate the

radical.

,

and 2.

Step 3 Solve the resulting equation.

Step 4 Check each proposed solution in the original

equation.






Example: Solve

Check:

− + =4 12 0 x x

= +4 12 x x

= +2

4 12 x x

− − =2

4 12 0 x x

( )− + =6 4 6 12 0

− =6 36 0

− =6 6 0

Solution: {6}

( )( )− + =6 2 0 x x

= −6, 2

=0 0

( )− − − + =2 4 2 12 0

− − =2 4 0

− − =2 2 0− ≠4 0




Example: Solve + − + =3 1 4 1 x




Example: Solve

Check:

+ − + =3 1 4 1 x

+ = + +3 1 4 1 x x

+ = + + + +3 1 4 2 4 1 x x x

− = +2 4 2 4 x x

( ) + − + =3 0 1 0 4 1

− =1 4 1

− =1 2 1

Solution: {5}

− = +2 4 x x

− + = +2

4 4 4 x x x

− =2

5 0 x x

( )− =5 0 x x

= 0, 5 x

− ≠

1 1( ) + − + =3 5 1 5 4 1

− =16 9 1

− =4 3 1=1 1



Quadratic in Form

An equation is said to be quadratic in form if it can be

written as

where a ≠ 0 and u is some algebraic expression.

+ + =2

0au bu c

To solve this type of equation, substitute u for the

algebraic expression, solve the quadratic expression for

u, and then set it equal to the algebraic expression and

solve for x . Because we are transforming the equation,you will still need to check any proposed solutions.



Quadratic in Form

Example: Solve ( ) ( )− + − − =2 3 1 3

1 1 12 0 x x



Quadratic in Form

Example: Solve

Let . This makes our equation:

( ) ( )− + − − =2 3 1 3

1 1 12 0 x x

( )= −

1 3

1u x

+ − =2

12 0u u

( )( )+ − =4 3 0u u

So, and= −4, 3u

( )− = −1 3

1 4 x ( )− =1 3

1 3 x

− = −1 64 x

= −63

− =1 27 x

= 28 x





Quadratic in Form

Example: Solve (cont.)( ) ( )− + − − =2 3 1 3

1 1 12 0 x x

( ) ( )− + − − =2 3 1 3

28 1 28 1 12 0

− =2 3 1 3

Solution: {–63, 28}

( ) ( )+ − =2 1

3 3 12 0

+ − =9 3 12 0

=0 0



Homework

College Algebra

Page 144: 35-85 (×5s) Turn In: 50, 55, 60, 80

Documents

3-4 Radical Equations (Presentation)