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
Math 17 (UP-IMath) Equations Lec 8 2 / 19
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Equations involving Rational Expressions
To solve:
Math 17 (UP-IMath) Equations Lec 8 3 / 19
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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).
Math 17 (UP-IMath) Equations Lec 8 3 / 19
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Example: Solve x+4x+2
2x3x5
= 3x8x23x10
.
Solution:
Math 17 (UP-IMath) Equations Lec 8 4 / 19
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Example: Solve x+4x+2
2x3x5
= 3x8x23x10
.
Solution:
x+4
x+2 2x3
x5 = 3x8
x23x10
Math 17 (UP-IMath) Equations Lec 8 4 / 19
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Example: Solve x+4x+2
2x3x5
= 3x8x23x10
.
Solution:
x+4
x+2 2x3
x5 = 3x8
x23x10
Math 17 (UP-IMath) Equations Lec 8 4 / 19
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Example: Solve x+4x+2
2x3x5
= 3x8x23x10
.
Solution:
(x 5)(x + 2)x+4
x+2 2x3
x5
= 3x8
x23x10
(x 5)(x + 2)
Math 17 (UP-IMath) Equations Lec 8 4 / 19
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Example: Solve x+4x+2
2x3x5
= 3x8x23x10
.
Solution:
(x 5)(x + 2)x+4
x+2 2x3
x5
= 3x8
x23x10
(x 5)(x + 2)= 3x 8
Math 17 (UP-IMath) Equations Lec 8 4 / 19
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Example: Solve x+4x+2
2x3x5
= 3x8x23x10
.
Solution:
(x 5)(x + 2) x+4x+2
2x3x5
= 3x8
x23x10
(x 5)(x + 2)(x 5)(x + 4) = 3x 8
Math 17 (UP-IMath) Equations Lec 8 4 / 19
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Example: Solve x+4x+2
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
Math 17 (UP-IMath) Equations Lec 8 4 / 19
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Example: Solve x+4x+2
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
Math 17 (UP-IMath) Equations Lec 8 4 / 19
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Example: Solve x+4x+2
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
Math 17 (UP-IMath) Equations Lec 8 4 / 19
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Example: Solve x+4x+2
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
Math 17 (UP-IMath) Equations Lec 8 4 / 19
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Example: Solve x+4x+2
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
Math 17 (UP-IMath) Equations Lec 8 4 / 19
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Example: Solve x+4x+2
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
Math 17 (UP-IMath) Equations Lec 8 4 / 19
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Example: Solve x+4x+2
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
Math 17 (UP-IMath) Equations Lec 8 4 / 19
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Example: Solve x+4x+2
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
Math 17 (UP-IMath) Equations Lec 8 4 / 19
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Example: Solve x+4x+2
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
Math 17 (UP-IMath) Equations Lec 8 4 / 19
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Example: Solve x+4x+2
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
Math 17 (UP-IMath) Equations Lec 8 4 / 19
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Example: Solve x+4x+2
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
Math 17 (UP-IMath) Equations Lec 8 4 / 19
4 2 3 3 8
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Example: Solve x+4x+2
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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Equations involving Radicals
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.
Math 17 (UP-IMath) Equations Lec 8 5 / 19
Equations involving Radicals
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Equations involving Radicals
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.
Example:
x + 1 = 2x = 1
Math 17 (UP-IMath) Equations Lec 8 5 / 19
Equations involving Radicals
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Equations involving Radicals
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.
Example:
x + 1 = 2x = 1
(x + 1)2 = 22
Math 17 (UP-IMath) Equations Lec 8 5 / 19
Equations involving Radicals
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Equations involving Radicals
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.
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
Math 17 (UP-IMath) Equations Lec 8 5 / 19
Equations involving Radicals
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Equations involving Radicals
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.
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
Math 17 (UP-IMath) Equations Lec 8 5 / 19
Equations in x involving Radicals
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Equations in x involving Radicals
Example: Solve for xin
x 2 = 2Solution:
x
2 = 2 Given
Math 17 (UP-IMath) Equations Lec 8 6 / 19
Equations in x involving Radicals
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Equations in x involving Radicals
Example: Solve for xin
x 2 = 2Solution:
x
2 = 2 Given
x 2 = 4 Square both sides
Math 17 (UP-IMath) Equations Lec 8 6 / 19
Equations in x involving Radicals
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Equations in x involving Radicals
Example: Solve for xin
x 2 = 2Solution:
x
2 = 2 Given
x 2 = 4 Square both sidesx = 6
Math 17 (UP-IMath) Equations Lec 8 6 / 19
Equations in x involving Radicals
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Equations in x involving Radicals
Example: Solve for xin
x 2 = 2Solution:
x
2 = 2 Given
x 2 = 4 Square both sidesx = 6
Checking: ifx= 6,
6 2 = 2
Math 17 (UP-IMath) Equations Lec 8 6 / 19
Equations in x involving Radicals
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q g
Example: Solve for xin
x 2 = 2Solution:
x
2 = 2 Given
x 2 = 4 Square both sidesx = 6
Checking: ifx= 6,
6 2 = 2
Hence, solution set is{6}.
Math 17 (UP-IMath) Equations Lec 8 6 / 19
Equations in x involving Radicals
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q g
Example: Solve for xin
x +
x 2 = 2Solution:
x +
x 2 = 2 Given
Math 17 (UP-IMath) Equations Lec 8 7 / 19
Equations in x involving Radicals
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q g
Example: Solve for xin
x +
x 2 = 2Solution:
x +
x 2 = 2 Givenx +
x 2 = 4 Square both sides
Math 17 (UP-IMath) Equations Lec 8 7 / 19
Equations in x involving Radicals
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q g
Example: Solve for xin
x +
x 2 = 2Solution:
x +
x 2 = 2 Givenx +
x 2 = 4 Square both sidesx
2 = 4
x Additive Property of Equality
Math 17 (UP-IMath) Equations Lec 8 7 / 19
Equations in x involving Radicals
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Example: Solve for xin
x +
x 2 = 2Solution:
x +
x 2 = 2 Givenx +
x 2 = 4 Square both sidesx
2 = 4
x Additive Property of Equality
x 2 = 16 8x + x2
Math 17 (UP-IMath) Equations Lec 8 7 / 19
Equations in x involving Radicals
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Example: Solve for xin
x +
x 2 = 2Solution:
x +
x 2 = 2 Givenx +
x 2 = 4 Square both sidesx
2 = 4
x Additive Property of Equality
x 2 = 16 8x + x2x2 9x + 18 = 0
Math 17 (UP-IMath) Equations Lec 8 7 / 19
Equations in x involving Radicals
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Example: Solve for xin
x +
x 2 = 2Solution:
x +
x 2 = 2 Givenx +
x 2 = 4 Square both sidesx
2 = 4
x Additive Property of Equality
x 2 = 16 8x + x2x2 9x + 18 = 0
(x 6)(x 3) = 0 factoring
Math 17 (UP-IMath) Equations Lec 8 7 / 19
Equations in x involving Radicals
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Example: Solve for xin
x +
x 2 = 2
Solution:x +
x 2 = 2 Given
x +
x 2 = 4 Square both sidesx
2 = 4
x Additive Property of Equality
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
Math 17 (UP-IMath) Equations Lec 8 7 / 19
Equations in x involving Radicals
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Example: Solve for xin
x +
x 2 = 2
Solution:x +
x 2 = 2 Given
x +
x 2 = 4 Square both sidesx
2 = 4
x Additive Property of Equality
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
x= 6 or x= 3
Math 17 (UP-IMath) Equations Lec 8 7 / 19
Equations in x involving Radicals
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Example: Solve for xin
x +
x 2 = 2
Solution:x +
x 2 = 2 Given
x +
x 2 = 4 Square both sidesx
2 = 4
x Additive Property of Equality
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
x= 6 or x= 3
Checking: ifx= 3,
3 + 3 2 = 2
ifx= 6,
6 +
6 2 = 8 = 22
Math 17 (UP-IMath) Equations Lec 8 7 / 19
Equations in x involving Radicals
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Example: Solve for xin
x +
x 2 = 2
Solution:x +
x 2 = 2 Given
x +
x 2 = 4 Square both sidesx
2 = 4
x Additive Property of Equality
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
x= 6 or x= 3
Checking: ifx= 3,
3 + 3 2 = 2
ifx= 6,
6 +
6 2 = 8 = 22= 2
Math 17 (UP-IMath) Equations Lec 8 7 / 19
Equations in x involving Radicals
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Example: Solve for xin
x +
x 2 = 2
Solution:x +
x 2 = 2 Given
x +
x 2 = 4 Square both sidesx
2 = 4
x Additive Property of Equality
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
x= 6 or x= 3
Checking: ifx= 3,
3 + 3 2 = 2
ifx= 6,
6 +
6 2 = 8 = 22= 2Hence, solution set is{3}.
Math 17 (UP-IMath) Equations Lec 8 7 / 19
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.
Math 17 (UP-IMath) Equations Lec 8 8 / 19
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 .
Math 17 (UP-IMath) Equations Lec 8 8 / 19
Solving Equations in Quadratic Form
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Example: Solve for xin 2
x + 1x
2
= 10 x 1x
.
Math 17 (UP-IMath) Equations Lec 8 9 / 19
Solving Equations in Quadratic Form
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Example: Solve for xin 2
x + 1x
2 = 10 x 1
x.
Solution:Equation is equivalent to 2
x + 1
x
2+
x + 1x
10 = 0.
Math 17 (UP-IMath) Equations Lec 8 9 / 19
Solving Equations in Quadratic Form
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Example: Solve for xin 2
x + 1x
2 = 10 x 1
x.
Solution:Equation is equivalent to 2
x + 1
x
2+
x + 1x
10 = 0.If=x + 1x , equation has the form 22 + 10 = 0.
Math 17 (UP-IMath) Equations Lec 8 9 / 19
Solving Equations in Quadratic Form
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Example: Solve for xin 2
x + 1x
2 = 10 x 1
x.
Solution:Equation is equivalent to 2
x + 1
x
2+
x + 1x
10 = 0.If=x + 1x , equation has the form 22 + 10 = 0.
22 + 10 = 0 quadratic form
Math 17 (UP-IMath) Equations Lec 8 9 / 19
Solving Equations in Quadratic Form
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Example: Solve for xin 2
x + 1x
2 = 10 x 1
x.
Solution:Equation is equivalent to 2
x + 1
x
2+
x + 1x
10 = 0.If=x + 1x , equation has the form 22 + 10 = 0.
22 + 10 = 0 quadratic form(2+ 5)( 2) = 0 factoring
Math 17 (UP-IMath) Equations Lec 8 9 / 19
Solving Equations in Quadratic Form
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Example: Solve for xin 2
x + 1x
2 = 10 x 1
x.
Solution:Equation is equivalent to 2
x + 1
x
2+
x + 1x
10 = 0.If=x + 1x , equation has the form 22 + 10 = 0.
22 + 10 = 0 quadratic form(2+ 5)( 2) = 0 factoring
=
5
2 or = 2
Math 17 (UP-IMath) Equations Lec 8 9 / 19
Solving Equations in Quadratic Form
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Example: Solve for xin 2
x + 1x
2 = 10 x 1
x.
Solution:Equation is equivalent to 2
x + 1
x
2+
x + 1x
10 = 0.If=x + 1x , equation has the form 22 + 10 = 0.
22 + 10 = 0 quadratic form(2+ 5)( 2) = 0 factoring
=
5
2 or = 2
To solve for x, we substitute back values for .
Math 17 (UP-IMath) Equations Lec 8 9 / 19
Solving Equations in Quadratic Form
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Example: Solve for xin 2
x + 1x
2 = 10 x 1
x.
Solution (cont):
=
5
2
or = 2
Math 17 (UP-IMath) Equations Lec 8 10 / 19
Solving Equations in Quadratic Form
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Example: Solve for xin 2
x + 1x
2 = 10 x 1
x.
Solution (cont):
=
5
2
or = 2
x + 1x
=
52
Math 17 (UP-IMath) Equations Lec 8 10 / 19
Solving Equations in Quadratic Form
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Example: Solve for xin 2
x + 1x
2 = 10 x 1
x.
Solution (cont):
=
5
2
or = 2
2xx + 1x
= 2x5
2
2x2 + 2 = 5x
Math 17 (UP-IMath) Equations Lec 8 10 / 19
Solving Equations in Quadratic Form
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Example: Solve for xin 2
x + 1x
2 = 10 x 1
x.
Solution (cont):
=
5
2
or = 2
2xx + 1x
= 2x5
2
2x2 + 2 = 5x
2x2 + 5x + 2 = 0
Math 17 (UP-IMath) Equations Lec 8 10 / 19
Solving Equations in Quadratic Form
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Example: Solve for xin 2
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
Math 17 (UP-IMath) Equations Lec 8 10 / 19
Solving Equations in Quadratic Form
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Example: Solve for xin 2
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]
Math 17 (UP-IMath) Equations Lec 8 10 / 19
Solving Equations in Quadratic Form
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Example: Solve for xin 2
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
Math 17 (UP-IMath) Equations Lec 8 10 / 19
Solving Equations in Quadratic Form
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Example: Solve for xin 2
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
Math 17 (UP-IMath) Equations Lec 8 10 / 19
Solving Equations in Quadratic Form
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Example: Solve for xin 2
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
Math 17 (UP-IMath) Equations Lec 8 10 / 19
Solving Equations in Quadratic Form
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Example: Solve for xin 2
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.
Math 17 (UP-IMath) Equations Lec 8 11 / 19
Solving Equations in Quadratic Form
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Example: Solve for xin 2
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
}.
Math 17 (UP-IMath) Equations Lec 8 11 / 19
Example: Solve for xin 6 + 3x + 1 = (x + 1) 23 .
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Math 17 (UP-IMath) Equations Lec 8 12 / 19
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.
Math 17 (UP-IMath) Equations Lec 8 12 / 19
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.If= (x + 1)
1
3 , equation has the form 2 6 = 0.
Math 17 (UP-IMath) Equations Lec 8 12 / 19
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.If= (x + 1)
1
3 , equation has the form 2 6 = 0.2 6 = 0 = 0 quadratic form
Math 17 (UP-IMath) Equations Lec 8 12 / 19
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.If= (x + 1)
1
3 , equation has the form 2 6 = 0.2 6 = 0 = 0 quadratic form( 3)(+ 2) = 0 factoring
Math 17 (UP-IMath) Equations Lec 8 12 / 19
Example: Solve for xin 6 + 3x + 1 = (x + 1) 23 .Solution:
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SEquation is equivalent to (x + 1)
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
= 3 or =2
Math 17 (UP-IMath) Equations Lec 8 12 / 19
Example: Solve for xin 6 + 3x + 1 = (x + 1) 23 .Solution:
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Equation is equivalent to (x + 1)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
= 3 or =2(x + 1) 13 = 3
Math 17 (UP-IMath) Equations Lec 8 12 / 19
Example: Solve for xin 6 + 3x + 1 = (x + 1) 23 .Solution:
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Equation is equivalent to (x + 1)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
= 3 or =2(x + 1) 13 = 3
(x + 1)1
3
3= (3)3
Math 17 (UP-IMath) Equations Lec 8 12 / 19
Example: Solve for xin 6 + 3x + 1 = (x + 1) 23 .Solution:
2 1
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Equation is equivalent to (x + 1)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
= 3 or =2(x + 1)
1
3 = 3(x + 1)
1
3
3= (3)3
x + 1 = 27
Math 17 (UP-IMath) Equations Lec 8 12 / 19
Example: Solve for xin 6 + 3x + 1 = (x + 1) 23 .Solution:
2 1
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Equation is equivalent to (x + 1)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
= 3 or =2(x + 1)
1
3 = 3(x + 1)
1
3
3= (3)3
x + 1 = 27x = 26
Math 17 (UP-IMath) Equations Lec 8 12 / 19
Example: Solve for xin 6 + 3x + 1 = (x + 1) 23 .Solution:
2 1
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Equation is equivalent to (x + 1)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
= 3 or =2(x + 1)
1
3 = 3(x + 1)
1
3
3= (3)3
x + 1 = 27x = 26
(x + 1)1
3 = 2
Math 17 (UP-IMath) Equations Lec 8 12 / 19
Example: Solve for xin 6 + 3x + 1 = (x + 1) 23 .Solution:
2 1
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Equation is equivalent to (x + 1)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
= 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
Math 17 (UP-IMath) Equations Lec 8 12 / 19
Example: Solve for xin 6 + 3x + 1 = (x + 1) 23 .Solution:
2 1
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Equation is equivalent to (x + 1)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
= 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
Math 17 (UP-IMath) Equations Lec 8 12 / 19
Example: Solve for xin 6 + 3x + 1 = (x + 1) 23 .Solution:
( )2
( )1
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Equation is equivalent to (x + 1)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
= 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
Math 17 (UP-IMath) Equations Lec 8 12 / 19
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.
Math 17 (UP-IMath) Equations Lec 8 12 / 19
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}.
Math 17 (UP-IMath) Equations Lec 8 12 / 19
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.
Math 17 (UP-IMath) Equations Lec 8 14 / 19
Equations involving Absolute Value
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Example: Solve x
2x3 = 2.
Solution:
Math 17 (UP-IMath) Equations Lec 8 14 / 19
Equations involving Absolute Value
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Example: Solve x
2x3 = 2.
Solution:
x
2x3 = 2 x
2x3 = 2
Math 17 (UP-IMath) Equations Lec 8 14 / 19
Equations involving Absolute Value
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Example: Solve x
2x3 = 2.
Solution:
x
2x3 = 2 x
2x3 = 2
Math 17 (UP-IMath) Equations Lec 8 14 / 19
Equations involving Absolute Value
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Example: Solve x
2x3 = 2.
Solution:
(2x 3)
x
2x3
= [2] (2x 3) x
2x3 = 2
Math 17 (UP-IMath) Equations Lec 8 14 / 19
Equations involving Absolute Value
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Example: Solve x
2x3 = 2.
Solution:
(2x 3)
x
2x3
= [2] (2x 3)
x = 4x 6
x
2x3 = 2
Math 17 (UP-IMath) Equations Lec 8 14 / 19
Equations involving Absolute Value
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Example: Solve x
2x3 = 2.
Solution:
(2x 3)
x
2x3
= [2] (2x 3)
x = 4x 66 = 3x
x
2x3 = 2
Math 17 (UP-IMath) Equations Lec 8 14 / 19
Equations involving Absolute Value
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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
Math 17 (UP-IMath) Equations Lec 8 14 / 19
Equations involving Absolute Value
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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
Math 17 (UP-IMath) Equations Lec 8 14 / 19
Equations involving Absolute Value
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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
Equations involving Absolute Value
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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
Math 17 (UP IMath) Equations Lec 8 14 / 19
Equations involving Absolute Value
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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
Math 17 (UP IMath) Equations Lec 8 14 / 19
Equations involving Absolute Value
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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
Math 17 (UP IMath) Equations Lec 8 14 / 19
Equations involving Absolute Value
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}
Math 17 (UP IMath) Equations Lec 8 14 / 19
Equations involving Absolute Value
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}
Math 17 (UP IMath) Equations Lec 8 14 / 19
Equations Involving Absolute Value
Theorem
Let R Then 2 2
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Leta
R. Then
|a
|2 =a2.
Example: Solve for xin|4x 3|=|x + 6|.
Math 17 (UP IMath) Equations Lec 8 15 / 19
Equations Involving Absolute Value
Theorem
Let a R Then a 2 a2
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Leta
R. Then
|a
|2 =a2.
Example: Solve for xin|4x 3|=|x + 6|.
Solution:
|4x 3|= |x + 6| Given
M th 17 (UP IM th) E ti s L 8 15 / 19
Equations Involving Absolute Value
Theorem
Let a R Then a 2 = a2
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Leta
R. Then
|a
|=a .
Example: Solve for xin|4x 3|=|x + 6|.
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
Equations Involving Absolute Value
Theorem
Let a R Then a 2 = a2
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Leta
R. Then
|a
|=a .
Example: Solve for xin|4x 3|=|x + 6|.
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
Equations Involving Absolute Value
Theorem
Let a R Then a 2 = a2
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Leta
R. Then
|a
|=a .
Example: Solve for xin|4x 3|=|x + 6|.
Solution:
|4x 3|= |x + 6| Given|4x 3|2 = |x + 6|2 Square both sides(4x 3)2 = (x + 6)2 Theorem above
(4x 3)2 (x + 6)2 = 0 Additive Property
M th 17 (UP IM th) E ti L 8 15 / 19
Equations Involving Absolute Value
Theorem
Let a R Then a 2 = a2
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Leta
R. Then
|a
|=a .
Example: Solve for xin|4x 3|=|x + 6|.
Solution:
|4x 3|= |x + 6| Given|4x 3|2 = |x + 6|2 Square both sides(4x 3)2 = (x + 6)2 Theorem above
(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
Equations Involving Absolute Value
Theorem
Let a R. Then a 2 = a2.
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Leta
R. Then
|a
|a .
Example: Solve for xin|4x 3|=|x + 6|.
Solution:
|4x 3|= |x + 6| Given|4x 3|2 = |x + 6|2 Square both sides(4x 3)2 = (x + 6)2 Theorem above
(4x 3)2 (x + 6)2 = 0 Additive Property((4x
3) + (x+ 6))((4x
3)
(x+ 6)) = 0 Factor the LHS
x=35 or x= 3 Equate factors to0
M h 17 (UP IM h) E i L 8 15 / 19
Equations Involving Absolute Value
Theorem
Let a R. Then a 2 = a2.
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Leta
R. Then
|a
|a .
Example: Solve for xin|4x 3|=|x + 6|.
Solution:
|4x 3|= |x + 6| Given|4x 3|2 = |x + 6|2 Square both sides(4x 3)2 = (x + 6)2 Theorem above
(4x 3)2 (x + 6)2 = 0 Additive Property((4x
3) + (x+ 6))((4x
3)
(x+ 6)) = 0 Factor the LHS
x=35 or x= 3 Equate factors to0
After checking, solution set is{35
, 3}.
Math 17 (UP-IMath) Equations Lec 8 15 / 19
Equations Involving Absolute Value
Theorem
Leta R. Then a 2 =a2.
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| |Example: Solve for xin|4x 3|=x.
Math 17 (UP-IMath) Equations Lec 8 16 / 19
Equations Involving Absolute Value
Theorem
Leta R. Then|a|2 =a2.
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| |
Example: Solve for xin|4x 3|=x.
Solution:
|4x 3|= x Given
Math 17 (UP-IMath) Equations Lec 8 16 / 19
Equations Involving Absolute Value
Theorem
Leta R. Then|a|2 =a2.
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| |
Example: Solve for xin|4x 3|=x.
Solution:
|4x 3|= x Given|4x 3|2 = x2 square both sides
Math 17 (UP-IMath) Equations Lec 8 16 / 19
Equations Involving Absolute Value
Theorem
Leta R. Then|a|2 =a2.
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| |
Example: Solve for xin|4x 3|=x.
Solution:
|4x 3|= x Given|4x 3|2 = x2 square both sides(4x 3)2 = x2 theorem above
Math 17 (UP-IMath) Equations Lec 8 16 / 19
Equations Involving Absolute Value
Theorem
Leta R. Then|a|2 =a2.
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Example: Solve for xin|4x 3|=x.
Solution:
|4x 3|= x Given|4x 3|2 = x2 square both sides(4x 3)2 = x2 theorem above
(4x 3)2 x2 = 0 additive property
Math 17 (UP-IMath) Equations Lec 8 16 / 19
Equations Involving Absolute Value
Theorem
Leta R. Then|a|2 =a2.
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Example: Solve for xin|4x 3|=x.
Solution:
|4x 3|= x Given|4x 3|2 = x2 square both sides(4x 3)2 = x2 theorem above
(4x 3)2 x2 = 0 additive property((4x
3) + x)((4x
3)
x) = 0 diff. of squares
Math 17 (UP-IMath) Equations Lec 8 16 / 19
Equations Involving Absolute Value
Theorem
Leta R. Then|a|2 =a2.
http://find/ -
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Example: Solve for xin|4x 3|=x.
Solution:
|4x 3|= x Given|4x 3|2 = x2 square both sides(4x 3)2 = x2 theorem above
(4x 3)2 x2 = 0 additive property((4x
3) + x)((4x
3)
x) = 0 diff. of squares
x= 35 or x= 1 equate factors to 0
Math 17 (UP-IMath) Equations Lec 8 16 / 19
Equations Involving Absolute Value
Theorem
Leta R. Then|a|2 =a2.
http://find/ -
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Example: Solve for xin|4x 3|=x.
Solution:
|4x 3|= x Given|4x 3|2 = x2 square both sides(4x 3)2 = x2 theorem above
(4x 3)2 x2 = 0 additive property((4x
3) + x)((4x
3)
x) = 0 diff. of squares
x= 35 or x= 1 equate factors to 0
After checking, solution set is{35
, 1}.
Math 17 (UP-IMath) Equations Lec 8 16 / 19
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.
Math 17 (UP-IMath) Equations Lec 8 17 / 19
Example: Solve for xin x4 81 = 0Solution:
x4 81 = 0 Given
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Math 17 (UP-IMath) Equations Lec 8 18 / 19
Example: Solve for xin x4 81 = 0Solution:
x4 81 = 0 Given
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Factor the LHS
Math 17 (UP-IMath) Equations Lec 8 18 / 19
Example: Solve for xin x4 81 = 0Solution:
x4 81 = 0 Givenx2 x2
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(x
9)(x
+ 9) = 0 Factor the LHS
Math 17 (UP-IMath) Equations Lec 8 18 / 19
Example: Solve for xin x4 81 = 0Solution:
x4 81 = 0 Givenx2 x2
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(x
9)(x
+ 9) = 0 Factor the LHS(x 3)(x + 3)(x2 + 9) = 0
Math 17 (UP-IMath) Equations Lec 8 18 / 19
Example: Solve for xin x4 81 = 0Solution:
x4 81 = 0 Givenx2 x2
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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:
Math 17 (UP-IMath) Equations Lec 8 18 / 19
Example: Solve for xin x4 81 = 0Solution:
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
then equate each factor to 0:
x + 3 = 0
x = 3
Math 17 (UP-IMath) Equations Lec 8 18 / 19
Example: Solve for xin x4 81 = 0Solution:
x4 81 = 0 Given(x2 9)(x2 + 9) = 0 Factor the LHS
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(
)( )(x 3)(x + 3)(x2 + 9) = 0
then equate each factor to 0:
x + 3 = 0
x = 3x
3 = 0
x = 3
Math 17 (UP-IMath) Equations Lec 8 18 / 19
Example: Solve for xin x4 81 = 0
Solution:
x4 81 = 0 Given(x2 9)(x2 + 9) = 0 Factor the LHS
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(
)( )(x 3)(x + 3)(x2 + 9) = 0
then equate each factor to 0:
x + 3 = 0
x = 3x
3 = 0
x = 3 x2 + 9 = 0x2 = 9
x = 9x= 3i or x=3i
Math 17 (UP-IMath) Equations Lec 8 18 / 19
Example: Solve for xin x4 81 = 0
Solution:
x4 81 = 0 Given(x2 9)(x2 + 9) = 0 Factor the LHS
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(
)( )(x 3)(x + 3)(x2 + 9) = 0
then equate each factor to 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
Math 17 (UP-IMath) Equations Lec 8 19 / 19
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