a8 - equations rational radical - joma.pdf

7/25/2019 A8 - equations rational radical - joma.pdf

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Equations involving Rational Expressions, Radicals,

Absolute Values, and those in Quadratic Form

Mathematics 17

Institute of Mathematics, University of the Philippines-Diliman

Lecture 8

Math 17 (UP-IMath) Equations Lec 8 1 / 19
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Outline

1 Rational Equations in x

2 Equations involving Radicals

3 Equations in Quadratic Form

4 Equations Involving Absolute Values

5 Solving Equations via Factoring

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Equations involving Rational Expressions

To solve:

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Equations involving Rational Expressions

To solve:

1 Multiply the LCD to both sides of the equation

2 Solve the equation that arises

3 Drop any of the obtained solutions that make the rational expressionsundefined (extraneous solutions).

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Example: Solve x+4x+2

2x3x5

= 3x8x23x10

.

Solution:

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2x3x5

= 3x8x23x10

.

Solution:

x+4

x+2 2x3

x5 = 3x8

x23x10

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2x3x5

= 3x8x23x10

.

Solution:

x+4

x+2 2x3

x5 = 3x8

x23x10

http://find/http://goback/


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2x3x5

= 3x8x23x10

.

Solution:

(x 5)(x + 2)x+4

x+2 2x3

x5

= 3x8

x23x10

(x 5)(x + 2)



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2x3x5

= 3x8x23x10

.

Solution:

(x 5)(x + 2)x+4

x+2 2x3

x5

= 3x8

x23x10

(x 5)(x + 2)= 3x 8

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2x3x5

= 3x8x23x10

.

Solution:

(x 5)(x + 2) x+4x+2

2x3x5

= 3x8

x23x10

(x 5)(x + 2)(x 5)(x + 4) = 3x 8

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2x3x5

= 3x8x23x10

.

Solution:

(x 5)(x + 2) x+4x+2

2x3x5

= 3x8

x23x10

(x 5)(x + 2)(x 5)(x + 4) (2x 3)(x + 2) = 3x 8

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2x3x5

= 3x8x23x10

.

Solution:

(x 5)(x + 2) x+4x+2

2x3x5

= 3x8

x23x10

(x 5)(x + 2)(x 5)(x + 4) (2x 3)(x + 2) = 3x 8

= 3x 8

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2x3x5

= 3x8x23x10

.

Solution:

(x 5)(x + 2) x+4x+2

2x3x5

= 3x8

x23x10

(x 5)(x + 2)(x 5)(x + 4) (2x 3)(x + 2) = 3x 8

x2 x 20 = 3x 8



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2x3x5

= 3x8x23x10

.

Solution:

(x 5)(x + 2) x+4x+2

2x3x5

= 3x8

x23x10

(x 5)(x + 2)(x 5)(x + 4) (2x 3)(x + 2) = 3x 8

x2 x 20 (2x2 + x 6) = 3x 8

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2x3x5

= 3x8x23x10

.

Solution:

(x 5)(x + 2) x+4x+2

2x3x5

= 3x8

x23x10

(x 5)(x + 2)(x 5)(x + 4) (2x 3)(x + 2) = 3x 8

x2 x 20 (2x2 + x 6) = 3x 8x2 x 20 2x2 x + 6 = 3x 8

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2x3x5

= 3x8x23x10

.

Solution:

(x 5)(x + 2) x+4x+2

2x3x5

= 3x8

x23x10

(x 5)(x + 2)(x 5)(x + 4) (2x 3)(x + 2) = 3x 8

x2 x 20 (2x2 + x 6) = 3x 8x2 x 20 2x2 x + 6 = 3x 8

x2

5x 6 = 0

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2x3x5

= 3x8x23x10

.

Solution:

(x 5)(x + 2) x+4x+2

2x3x5

= 3x8

x23x10

(x 5)(x + 2)(x 5)(x + 4) (2x 3)(x + 2) = 3x 8

x2 x 20 (2x2 + x 6) = 3x 8x2 x 20 2x2 x + 6 = 3x 8

x2

5x 6 = 0x2 + 5x + 6 = 0

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2x3x5

= 3x8x23x10

.

Solution:

(x 5)(x + 2) x+4x+2

2x3x5

= 3x8

x23x10

(x 5)(x + 2)(x 5)(x + 4) (2x 3)(x + 2) = 3x 8

x2 x 20 (2x2 + x 6) = 3x 8x2 x 20 2x2 x + 6 = 3x 8

x2

5x 6 = 0x2 + 5x + 6 = 0

(x + 2)(x + 3) = 0

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2x3x5

= 3x8x23x10

.

Solution:

(x 5)(x + 2) x+4x+2

2x3x5

= 3x8

x23x10

(x 5)(x + 2)(x 5)(x + 4) (2x 3)(x + 2) = 3x 8

x2 x 20 (2x2 + x 6) = 3x 8x2 x 20 2x2 x + 6 = 3x 8

x2

5x 6 = 0x2 + 5x + 6 = 0

(x + 2)(x + 3) = 0x + 2 = 0 or x + 3 = 0

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2x3x5

= 3x8x23x10

.

Solution:

(x 5)(x + 2) x+4x+2

2x3x5

= 3x8

x23x10

(x 5)(x + 2)(x 5)(x + 4) (2x 3)(x + 2) = 3x 8

x2 x 20 (2x2 + x 6) = 3x 8x2 x 20 2x2 x + 6 = 3x 8

x2

5x 6 = 0x2 + 5x + 6 = 0

(x + 2)(x + 3) = 0x + 2 = 0 or x + 3 = 0

x=

2 or x=

3

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2x3x5

= 3x8x23x10

.

Solution:

(x 5)(x + 2) x+4x+2

2x3x5

= 3x8

x23x10

(x 5)(x + 2)(x 5)(x + 4) (2x 3)(x + 2) = 3x 8

x2 x 20 (2x2 + x 6) = 3x 8x2 x 20 2x2 x + 6 = 3x 8

x2

5x 6 = 0x2 + 5x + 6 = 0

(x + 2)(x + 3) = 0x + 2 = 0 or x + 3 = 0

x=

2 or x=

3


4 2 3 3 8
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2x3x5

= 3x8x23x10

.

Solution:

(x 5)(x + 2) x+4x+2

2x3x5

= 3x8

x23x10

(x 5)(x + 2)(x 5)(x + 4) (2x 3)(x + 2) = 3x 8

x2 x 20 (2x2 + x 6) = 3x 8x2 x 20 2x2 x + 6 = 3x 8

x2

5x 6 = 0x2 + 5x + 6 = 0

(x + 2)(x + 3) = 0x + 2 = 0 or x + 3 = 0

x=

2 or x=

3

Checking: Ifx=2, a denominator will be zero.Ifx=3, x+4

x+2 2x3

x5 =17

8 and 3x8

x23x10

=178

.

Hence, solution set is{3}.Math 17 (UP-IMath) Equations Lec 8 4 / 19

Equations involving Radicals


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TheoremLetA andB be expressions and letn N.Then the solution set of the equation A=B is a subset of the solution setof the equation An =Bn.


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Example:

x + 1 = 2x = 1




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Example:

x + 1 = 2x = 1

(x + 1)2 = 22


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Example:

x + 1 = 2x = 1

(x + 1)2 = 22

x2 + 2x + 1 = 4x2

+ 2x 3 = 0(x + 3)(x 1) = 0

x=3 or x= 1


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Example:

x + 1 = 2x = 1

(x + 1)2 = 22

x2 + 2x + 1 = 4x2

+ 2x 3 = 0(x + 3)(x 1) = 0

x=3 or x= 13 is an extraneous solution


Equations in x involving Radicals


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Example: Solve for xin

x 2 = 2Solution:

x

2 = 2 Given




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x 2 = 2Solution:

x

2 = 2 Given

x 2 = 4 Square both sides




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x 2 = 2Solution:

x

2 = 2 Given

x 2 = 4 Square both sidesx = 6


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x 2 = 2Solution:

x

2 = 2 Given


Checking: ifx= 6,

6 2 = 2




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q g


x 2 = 2Solution:

x

2 = 2 Given


Checking: ifx= 6,

6 2 = 2

Hence, solution set is{6}.


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q g


x +

x 2 = 2Solution:

x +

x 2 = 2 Given


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q g


x +

x 2 = 2Solution:

x +

x 2 = 2 Givenx +

x 2 = 4 Square both sides


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q g


x +

x 2 = 2Solution:

x +

x 2 = 2 Givenx +

x 2 = 4 Square both sidesx

2 = 4

x Additive Property of Equality


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x +

x 2 = 2Solution:

x +

x 2 = 2 Givenx +


2 = 4


x 2 = 16 8x + x2


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x +

x 2 = 2Solution:

x +

x 2 = 2 Givenx +


2 = 4


x 2 = 16 8x + x2x2 9x + 18 = 0


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x +

x 2 = 2Solution:

x +

x 2 = 2 Givenx +


2 = 4


x 2 = 16 8x + x2x2 9x + 18 = 0

(x 6)(x 3) = 0 factoring


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x +

x 2 = 2

Solution:x +

x 2 = 2 Given

x +


2 = 4


x 2 = 16 8x + x2x2 9x + 18 = 0

(x 6)(x 3) = 0 factoringx 6 = 0 or x 3 = 0 ab= 0a= 0 orb= 0


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x +

x 2 = 2

Solution:x +

x 2 = 2 Given

x +


2 = 4


x 2 = 16 8x + x2x2 9x + 18 = 0


x= 6 or x= 3


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x +

x 2 = 2

Solution:x +

x 2 = 2 Given

x +


2 = 4


x 2 = 16 8x + x2x2 9x + 18 = 0


x= 6 or x= 3

Checking: ifx= 3,

3 + 3 2 = 2

ifx= 6,

6 +

6 2 = 8 = 22


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x +

x 2 = 2

Solution:x +

x 2 = 2 Given

x +


2 = 4


x 2 = 16 8x + x2x2 9x + 18 = 0


x= 6 or x= 3

Checking: ifx= 3,

3 + 3 2 = 2

ifx= 6,

6 +

6 2 = 8 = 22= 2


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x +

x 2 = 2

Solution:x +

x 2 = 2 Given

x +


2 = 4


x 2 = 16 8x + x2x2 9x + 18 = 0


x= 6 or x= 3

Checking: ifx= 3,

3 + 3 2 = 2

ifx= 6,

6 +

6 2 = 8 = 22= 2Hence, solution set is{3}.


Equations in Quadratic Form
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Definition

An equation is inquadratic formif it can be written in the form

a2 + b+ c= 0,

where a, b, c R and a= 0.


Equations in Quadratic Form
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Definition

An equation is inquadratic formif it can be written in the form

a2 + b+ c= 0,

where a, b, c R and a= 0.

Example:

2

x + 1

x2

+

x + 1

x 10 = 0 is in quadratic form if=x +

1

x .


Solving Equations in Quadratic Form
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Example: Solve for xin 2

x + 1x

2

= 10 x 1x

.


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x + 1x

2 = 10 x 1

x.

Solution:Equation is equivalent to 2

x + 1

x

2+

x + 1x

10 = 0.


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x + 1x

2 = 10 x 1

x.


x + 1

x

2+

x + 1x

10 = 0.If=x + 1x , equation has the form 22 + 10 = 0.


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x + 1x

2 = 10 x 1

x.


x + 1

x

2+

x + 1x


22 + 10 = 0 quadratic form


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x + 1x

2 = 10 x 1

x.


x + 1

x

2+

x + 1x


22 + 10 = 0 quadratic form(2+ 5)( 2) = 0 factoring


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x + 1x

2 = 10 x 1

x.


x + 1

x

2+

x + 1x



=

5

2 or = 2


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x + 1x

2 = 10 x 1

x.


x + 1

x

2+

x + 1x



=

5

2 or = 2

To solve for x, we substitute back values for .


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x + 1x

2 = 10 x 1

x.

Solution (cont):

=

5

2

or = 2


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x + 1x

2 = 10 x 1

x.

Solution (cont):

=

5

2

or = 2

x + 1x

=

52


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x + 1x

2 = 10 x 1

x.

Solution (cont):

=

5

2

or = 2

2xx + 1x

= 2x5

2

2x2 + 2 = 5x


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x + 1x

2 = 10 x 1

x.

Solution (cont):

=

5

2

or = 2

2xx + 1x

= 2x5

2

2x2 + 2 = 5x

2x2 + 5x + 2 = 0


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x + 1x

2 = 10 x 1

x.

Solution (cont):

=

5

2

or = 2

2xx + 1x

= 2x5

2

2x2 + 2 = 5x

2x2 + 5x + 2 = 0(2x + 1)(x + 2) = 0

x=12 or x=2


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x + 1x

2 = 10 x 1

x.

Solution (cont):

=

5

2

or = 2

2xx + 1x

= 2x5

2

2x2 + 2 = 5x

2x2 + 5x + 2 = 0(2x + 1)(x + 2) = 0

x=12 or x=2

x + 1

x

= [2]


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x + 1x

2 = 10 x 1

x.

Solution (cont):

=

5

2

or = 2

2xx + 1x

= 2x5

2

2x2 + 2 = 5x

2x2 + 5x + 2 = 0(2x + 1)(x + 2) = 0

x=12 or x=2

xx + 1x

= x[2]

x2 + 1 = 2x


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x + 1x

2 = 10 x 1

x.

Solution (cont):

=

5

2

or = 2

2xx + 1x

= 2x5

2

2x2 + 2 = 5x

2x2 + 5x + 2 = 0(2x + 1)(x + 2) = 0

x=12 or x=2

xx + 1x

= x[2]

x2 + 1 = 2xx2 2x + 1 = 0


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x + 1x

2 = 10 x 1

x.

Solution (cont):

=

5

2

or = 2

2xx + 1x

= 2x5

2

2x2 + 2 = 5x

2x2 + 5x + 2 = 0(2x + 1)(x + 2) = 0

x=12 or x=2

xx + 1x

= x[2]

x2 + 1 = 2xx2 2x + 1 = 0

(x

1)2 = 0

x = 1


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x + 1x

2= 10 x 1

x.

Solution (cont):

Checking:if x= 12 , 2 x+ 1x

2

= 25

2 and 10 x 1x = 252 .if x= 2, 2 x+ 1

x

2

= 252

and 10 x 1x

= 252 .

if x= 1, 2x+ 1

x

2

= 8 and 10 x 1x

= 8.


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x + 1x

2= 10 x 1

x.

Solution (cont):

Checking:if x= 12 , 2 x+ 1x

2

= 25

2 and 10 x 1x = 252 .if x= 2, 2 x+ 1

x

2

= 252

and 10 x 1x

= 252 .

if x= 1, 2x+ 1

x

2

= 8 and 10 x 1x

= 8.

Hence, solution set is

{1

2,

2, 1

}.


Example: Solve for xin 6 + 3x + 1 = (x + 1) 23 .
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Example: Solve for xin 6 + 3x + 1 = (x + 1) 23 .Solution:
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Solution:Equation is equivalent to (x + 1)

2

3 (x + 1) 13 6 = 0.


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2

3 (x + 1) 13 6 = 0.If= (x + 1)

1

3 , equation has the form 2 6 = 0.


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2

3 (x + 1) 13 6 = 0.If= (x + 1)

1

3 , equation has the form 2 6 = 0.2 6 = 0 = 0 quadratic form


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2

3 (x + 1) 13 6 = 0.If= (x + 1)

1

3 , equation has the form 2 6 = 0.2 6 = 0 = 0 quadratic form( 3)(+ 2) = 0 factoring


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SEquation is equivalent to (x + 1)

2

3 (x + 1) 13 6 = 0.If= (x + 1)

1


= 3 or =2


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Equation is equivalent to (x + 1)2

3 (x + 1) 13 6 = 0.If= (x + 1)

1


= 3 or =2(x + 1) 13 = 3


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3 (x + 1) 13 6 = 0.

If= (x + 1)1


= 3 or =2(x + 1) 13 = 3

(x + 1)1

3

3= (3)3



2 1
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3 (x + 1) 13 6 = 0.

If= (x + 1)1


= 3 or =2(x + 1)

1

3 = 3(x + 1)

1

3

3= (3)3

x + 1 = 27



2 1
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3 (x + 1) 13 6 = 0.

If= (x + 1)1


= 3 or =2(x + 1)

1

3 = 3(x + 1)

1

3

3= (3)3

x + 1 = 27x = 26



2 1
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3 (x + 1) 13 6 = 0.

If= (x + 1)1


= 3 or =2(x + 1)

1

3 = 3(x + 1)

1

3

3= (3)3

x + 1 = 27x = 26

(x + 1)1

3 = 2



2 1
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3 (x + 1) 13 6 = 0.

If= (x + 1)1


= 3 or =2(x + 1)

1

3 = 3(x + 1)

1

3

3= (3)3

x + 1 = 27x = 26

(x + 1)1

3 = 2(x + 1)

1

3

3= (2)3



2 1
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3 (x + 1) 13 6 = 0.

If= (x + 1)1


2 6 = 0 = 0 quadratic form( 3)(+ 2) = 0 factoring

= 3 or =2(x + 1)

1

3 = 3(x + 1)

1

3

3= (3)3

x + 1 = 27x = 26

(x + 1)1

3 = 2(x + 1)

1

3

3= (2)3

x + 1 = 8



( )2

( )1
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3 (x + 1) 13 6 = 0.

If= (x + 1)1


2 6 = 0 = 0 quadratic form( 3)(+ 2) = 0 factoring

= 3 or =2(x + 1)

1

3 = 3(x + 1)

1

3

3= (3)3

x + 1 = 27x = 26

(x + 1)1

3 = 2(x + 1)

1

3

3= (2)3

x + 1 = 8x + 1 =

9


Example: Solve for xin 6 + 3x + 1 = (x + 1) 23 .Solution:E i i i l ( 1)

2

( 1)1

6 0
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Equation is equivalent to (x + 1) 3 (x + 1) 3 6 = 0.

If= (x + 1)1

3 , equation has the form 2

6 = 0.2 6 = 0 = 0 quadratic form( 3)(+ 2) = 0 factoring

= 3 or =2(x + 1)

1

3 = 3(x + 1)

1

3

3= (3)3

x + 1 = 27x = 26

(x + 1)1

3 = 2(x + 1)

1

3

3= (2)3

x + 1 = 8x + 1 =

9

Checking: if x= 26, 6 + 3

x+ 1 = 9 and (x+ 1)2

3 = 9.

if x= 9, 6 + 3x+ 1 = 4 and (x+ 1) 23 = 4.


Example: Solve for xin 6 + 3x + 1 = (x + 1) 23 .Solution:E i i i l ( + 1)

2

( + 1)1

6 0
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Equation is equivalent to (x + 1) 3 (x + 1) 3 6 = 0.

If= (x + 1)1

3 , equation has the form 2

6 = 0.2 6 = 0 = 0 quadratic form( 3)(+ 2) = 0 factoring

= 3 or =2(x + 1)

1

3 = 3(x + 1)

1

3

3= (3)3

x + 1 = 27x = 26

(x + 1)1

3 = 2(x + 1)

1

3

3= (2)3

x + 1 = 8x + 1 =

9

Checking: if x= 26, 6 + 3

x+ 1 = 9 and (x+ 1)2

3 = 9.

if x= 9, 6 + 3x+ 1 = 4 and (x+ 1) 23 = 4.Hence, solution set is{26,9}.


Equations Involving Absolute Value
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Let Ebe an expression and a R.Ifa >0, then|E|= a if and only ifE=a orE=a.Ifa= 0, then|E|= a if and only ifE= 0.Ifa


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Example: Solve x

2x3 = 2.


Equations involving Absolute Value
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Example: Solve x

2x3 = 2.

Solution:


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Example: Solve x

2x3 = 2.

Solution:

x

2x3 = 2 x

2x3 = 2


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Example: Solve x

2x3 = 2.

Solution:

x

2x3 = 2 x

2x3 = 2


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Example: Solve x

2x3 = 2.

Solution:

(2x 3)

x

2x3

= [2] (2x 3) x

2x3 = 2


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Example: Solve x

2x3 = 2.

Solution:

(2x 3)

x

2x3

= [2] (2x 3)

x = 4x 6

x

2x3 = 2


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Example: Solve x

2x3 = 2.

Solution:

(2x 3)

x

2x3

= [2] (2x 3)

x = 4x 66 = 3x

x

2x3 = 2


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Example: Solve x

2x3 = 2.

Solution:

(2x 3)

x

2x3

= [2] (2x 3)

x = 4x 66 = 3x2 = x

x

2x3 = 2


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Example: Solve

x

2x3

= 2.

Solution:

(2x 3)

x

2x3

= [2] (2x 3)

x = 4x 66 = 3x2 = x

x

2x3 = 2


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Example: Solve

x

2x3

= 2.

Solution:

(2x 3)

x

2x3

= [2] (2x 3)

x = 4x 66 = 3x2 = x

(2x 3)

x

2x3

= [ 2] (2x 3)

Math 17 (UP IMath) Equations Lec 8 14 / 19

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Example: Solve

x

2x3

= 2.

Solution:

(2x 3)

x

2x3

= [2] (2x 3)

x = 4x 66 = 3x2 = x

(2x 3)

x

2x3

= [ 2] (2x 3)

x = 4x + 6


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Example: Solve

x

2x3

= 2.

Solution:

(2x 3)

x

2x3

= [2] (2x 3)

x = 4

x 66 = 3x

2 = x

(2x 3)

x

2x3

= [ 2] (2x 3)

x = 4

x+ 65x = 6


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Example: Solve

x

2x3

= 2.

Solution:

(2x 3)

x

2x3

= [2] (2x 3)

x = 4

x 66 = 3x

2 = x

(2x 3)

x

2x3

= [ 2] (2x 3)

x = 4

x+ 65x = 6

x = 65



S

x
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Example: Solve

x

2x3

= 2.

Solution:

(2x 3)

x

2x3

= [2] (2x 3)

x = 4

x 66 = 3x

2 = x

(2x 3)

x

2x3

= [ 2](2x 3)

x = 4

x+ 65x = 6

x = 65

Checking: if x= 2, | x

2x3| = 2.

if x= 65, | x

2x3| = 2.

Solution set is{65

, 2}



E l S l

x

2


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Example: Solve

x

2x3

= 2.

Solution:

x

2x3 = 2

x = 4x

66 = 3x2 = x

x

2x3 = 2

x =

4x + 65x = 6

x = 65

Checking: if x= 2, | x

2x3| = 2.

if x= 65, | x

2x3| = 2.

Solution set is{65

, 2}



Theorem

Let R Then 2 2
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Leta

R. Then

|a

|2 =a2.

Example: Solve for xin|4x 3|=|x + 6|.



Theorem

Let a R Then a 2 a2
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Leta

R. Then

|a

|2 =a2.


Solution:

|4x 3|= |x + 6| Given

M th 17 (UP IM th) E ti s L 8 15 / 19


Theorem

Let a R Then a 2 = a2
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Leta

R. Then

|a

|=a .


Solution:

|4x 3|= |x + 6| Given|4x 3|2 = |x + 6|2 Square both sides

M th 17 (UP IM th) E ti L 8 15 / 19


Theorem

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Leta

R. Then

|a

|=a .


Solution:

|4x 3|= |x + 6| Given|4x 3|2 = |x + 6|2 Square both sides(4x 3)2 = (x + 6)2 Theorem above

M th 17 (UP IM th) E ti L 8 15 / 19


Theorem

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Leta

R. Then

|a

|=a .


Solution:


(4x 3)2 (x + 6)2 = 0 Additive Property

M th 17 (UP IM th) E ti L 8 15 / 19


Theorem

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Leta

R. Then

|a

|=a .


Solution:


(4x 3)2 (x + 6)2 = 0 Additive Property((4x

3) + (x+ 6))((4x

3)

(x+ 6)) = 0 Factor the LHS

M h 17 (UP IM h) E i L 8 15 / 19


Theorem

Let a R. Then a 2 = a2.
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Leta

R. Then

|a

|a .


Solution:



3) + (x+ 6))((4x

3)


x=35 or x= 3 Equate factors to0

M h 17 (UP IM h) E i L 8 15 / 19


Theorem

Let a R. Then a 2 = a2.
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Leta

R. Then

|a

|a .


Solution:



3) + (x+ 6))((4x

3)


x=35 or x= 3 Equate factors to0

After checking, solution set is{35

, 3}.



Theorem

Leta R. Then a 2 =a2.
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| |Example: Solve for xin|4x 3|=x.



Theorem

Leta R. Then|a|2 =a2.
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| |

Example: Solve for xin|4x 3|=x.

Solution:

|4x 3|= x Given



Theorem

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| |


Solution:

|4x 3|= x Given|4x 3|2 = x2 square both sides



Theorem

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| |


Solution:

|4x 3|= x Given|4x 3|2 = x2 square both sides(4x 3)2 = x2 theorem above



Theorem

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Solution:


(4x 3)2 x2 = 0 additive property



Theorem

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Solution:


(4x 3)2 x2 = 0 additive property((4x

3) + x)((4x

3)

x) = 0 diff. of squares



Theorem

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Solution:



3) + x)((4x

3)


x= 35 or x= 1 equate factors to 0



Theorem

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Solution:



3) + x)((4x

3)


x= 35 or x= 1 equate factors to 0

After checking, solution set is{35

, 1}.


Solving Equations via Factoring
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TheoremFor allx1, x2,...,xn R,

x1 x2 ... xn= 0 if and only if x1= 0 orx2= 0 or ... orxn= 0

Let Ebe an expression. For equations with the form E= 0, we try tofactor Eand equate each of its factors to 0.

Note: This only works when one side of the equation is 0.


Example: Solve for xin x4 81 = 0Solution:

x4 81 = 0 Given
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x4 81 = 0 Given
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Factor the LHS



x4 81 = 0 Givenx2 x2
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(x

9)(x

+ 9) = 0 Factor the LHS



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(x

9)(x

+ 9) = 0 Factor the LHS(x 3)(x + 3)(x2 + 9) = 0



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(x

9)(x

+ 9) = 0 Factor the LHS(x 3)(x + 3)(x2 + 9) = 0

then equate each factor to 0:



x4 81 = 0 Given(x2 9)(x2 + 9) = 0 Factor the LHS


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(

9)( + 9) 0 S(x 3)(x + 3)(x2 + 9) = 0


x + 3 = 0

x = 3



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(

)( )(x 3)(x + 3)(x2 + 9) = 0


x + 3 = 0

x = 3x

3 = 0

x = 3


Example: Solve for xin x4 81 = 0

Solution:

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(

)( )(x 3)(x + 3)(x2 + 9) = 0


x + 3 = 0

x = 3x

3 = 0

x = 3 x2 + 9 = 0x2 = 9

x = 9x= 3i or x=3i


Example: Solve for xin x4 81 = 0

Solution:

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(

)( )(x 3)(x + 3)(x2 + 9) = 0


x + 3 = 0

x = 3x

3 = 0

x = 3 x2 + 9 = 0x2 = 9

x = 9x= 3i or x=3i

Checking: if x= 3, x4

81 = 0 ifx= 3, x4

81 = 0if x= 3i,x4 81 = 0 ifx= 3i, x4 81 = 0

Solution set is{3,3, 3i,3i}Math 17 (UP-IMath) Equations Lec 8 18 / 19

Exercises:
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Solve for x:1 5

x+3= 8+x

x+3 3

2 x

x1+ x5

x2+2x3

= 1x+3

3

2x + 3 x 2 = x + 14 x6 = 8 7x35

3x42x+3

= 1

Next Meeting: Inequalities

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a8 - equations rational radical - joma.pdf

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