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Section A.6
Solving Equations
Math 1051 - Precalculus I
Solving Equations
Section A.6
A.6 Solving Equations
Simplify
1
x +2
−
2x − 6 x 2 − 9
Solving Equations
Section A.6
A.6 Solving Equations
Simplify
x +2
1
− 2
2x − 6 x − 9
Factor:
2x − 6 = 2(x − 3)
2
x − 9 = (x − 3)(x + 3)
LCD = 2(x − 3)(x + 3)
Solving Equations
Section A.6
A.6 Solving Equations
Simplify
1
x +2
−
2x − 6 x 2 − 9
LCD = 2(x − 3)(x + 3), so
1
x +2
− 2
2x − 6 x − 9
=
=
=
1
x +3
x +2
2
·
−
·
2(x − 3) x + 3 (x − 3)(x + 3) 2
(x + 3) − (2x + 4)
2(x − 3)(x + 3)
−x − 1
2(x − 3)(x + 3)
Solving Equations
Section A.6
Solving Equations
Equation: Two expressions set equal to each other.
Solving Equations
Section A.6
Solving Equations
Equation: Two expressions set equal to each other.
To solve means to find the values of the variables that make the
equation a true statement.
Solving Equations
Section A.6
Solving Equations
Equation: Two expressions set equal to each other.
To solve means to find the values of the variables that make the
equation a true statement.
A solution must be in the domain of the original equation.
Solving Equations
Section A.6
Three types of solutions
Solving Equations
Section A.6
Three types of solutions
Conditional: True for some values of x but not others.
x + 2 = 3 is a conditional equation because 1 is a solution but
other numbers are not.
Solving Equations
Section A.6
Three types of solutions
Conditional: True for some values of x but not others.
x + 2 = 3 is a conditional equation because 1 is a solution but
other numbers are not.
Contradiction: An equation that is false for all values of x.
There is no solution.
x + 1 = x is a contradiction. If you take a number and add 1 to
it, you don’t get back the original number.
Solving Equations
Section A.6
Three types of solutions
Conditional: True for some values of x but not others.
x + 2 = 3 is a conditional equation because 1 is a solution but
other numbers are not.
Contradiction: An equation that is false for all values of x.
There is no solution.
x + 1 = x is a contradiction. If you take a number and add 1 to
it, you don’t get back the original number.
Identity: An equation that is true for all values of x. The solution
is all real numbers.
x + 1 = 1 + x is an identity. It is an example of the commutative
property of addition.
Solving Equations
Section A.6
−2
1
(2x + 3) = (3 − x) − 1
3
2
Solving Equations
Section A.6
−2
1
(2x + 3) = (3 − x) − 1
3
2
5
-6
-4
-2
2
-5
-10
Solving Equations
Section A.6
4
6
5(x − 4) − (3 − x) = 2(x + 5) + 4x
Solving Equations
Section A.6
5(x − 4) − (3 − x) = 2(x + 5) + 4x
6
4
2
-6
-4
-2
2
-2
-4
-6
Solving Equations
Section A.6
4
6
5(x − 4) − (3 − x) = 2(x + 5) + 4x
6
4
2
-6
-4
-2
2
-2
-4
-6
No solution!
Solving Equations
Section A.6
4
6
6x 2 − 5 = 13x
Solving Equations
Section A.6
6x 2 − 5 = 13x
40
30
20
10
-4
-2
Solving Equations
2
Section A.6
4
6x 2 − 5 = 13x
40
30
20
10
-4
-2
2
We can check that both of these work
Solving Equations
Section A.6
4
Factoring
3x 2 + 30x = 6x 3
Solving Equations
Section A.6
Factoring
3x 2 + 30x = 6x 3
100
50
-4
-2
2
-50
Solving Equations
Section A.6
4
Absolute Values
4 |3x − 4| = 20
Solving Equations
Section A.6
Absolute Values
4 |3x − 4| = 20
Rule for absolute values
If |y | = b, then y = b or y = −b. Use this to split the equation
into 2 different equations without an absolute value.
Solving Equations
Section A.6
Absolute Values
4 |3x − 4| = 20
30
25
20
15
10
5
-6
-4
-2
Solving Equations
2
Section A.6
4
6
5 x 2 + 3x − 2 − 3 = 7
Solving Equations
Section A.6
5 x 2 + 3x − 2 − 3 = 7
20
15
10
5
-4
-2
2
-5
Solving Equations
Section A.6
4
Square Roots
(x − 2)2 = 9
Solving Equations
Section A.6
Square Roots
(x − 2)2 = 9
Rule for square roots
√
√
If y 2 = b, then y = b or y = − b. Use this to split the
equation into 2 different equations.
Solving Equations
Section A.6
Square Roots
(x − 2)2 = 9
10
8
6
4
2
-6
-4
-2
2
-2
-4
Solving Equations
Section A.6
4
6
Completing the Square
x 2 + 6x = 4
Solving Equations
Section A.6
Completing the Square
x 2 + 6x = 4
Completing the square
Since (x + a)2 = x 2 + 2ax + a2 , let b = 2a so that
quadratic equation like
x 2 + bx = c
2
we can solve by adding b2 to both sides.
Solving Equations
Section A.6
b
2
= a. In a
Completing the Square
2x 2 − 3x = 1
Solving Equations
Section A.6
Completing the Square
2x 2 − 3x = 1
10
8
6
4
2
-4
-2
2
-2
-4
Solving Equations
Section A.6
4
Quadratic Equation
The standard form for a quadratic equation is
ax 2 + bx + c = 0
where a, b, c are real numbers and a 6= 0.
We can use completing the square to solve this equation to get
the quadratic formula:
√
−b ± b2 − 4ac
x1,2 =
2a
Solving Equations
Section A.6
Quadratic Equation
2x 2 − 3x = 1
Solving Equations
Section A.6
Quadratic Equation
2x 2 − 3x = 1
Same answer as we found completing the square
Solving Equations
Section A.6
How about the cubic equation ax 3 + bx 2 + cx + d = 0? The
solution is:
Solving Equations
Section A.6
x1
=
−
−
x2
=
+
+
x3
=
+
+
−
b
3a
s
q
1 3 1
2b3 − 9abc + 27a2 d + (2b3 − 9abc + 27a2 d)2 − 4(b2 − 3ac)3
3a
2
s q
1 3 1
2b3 − 9abc + 27a2 d − (2b3 − 9abc + 27a2 d)2 − 4(b2 − 3ac)3
3a
2
−
b
3a
1+i
√ s 3 3 1
q
2b3 − 9abc + 27a2 d + (2b3 − 9abc + 27a2 d)2 − 4(b2 − 3ac)3
6a
2
√ s q
1−i 3 3 1
2b3 − 9abc + 27a2 d − (2b3 − 9abc + 27a2 d)2 − 4(b2 − 3ac)3
6a
2
−
b
3a
1−i
√ s 3 3 1
6a
2
√ s 1+i 3 3 1
6a
2
2b3 − 9abc + 27a2 d +
q
(2b3 − 9abc + 27a2 d)2 − 4(b2 − 3ac)3
2b3 − 9abc + 27a2 d −
q
(2b3 − 9abc + 27a2 d)2 − 4(b2 − 3ac)3
Solving Equations
Section A.6
Gerolamo Cardano
Solving Equations
Section A.6
Cubic Equations
No magical formula. Hopefully we can factor.
x3 − 1 = 0
Solving Equations
Section A.6
Cubic Equations
No magical formula. Hopefully we can factor.
x3 − 1 = 0
10
5
-4
-2
2
-5
-10
Solving Equations
Section A.6
4
Solve:
2x 2 − x
−x 2 − 2x = (2x − 1)x
Solving Equations
Section A.6
Solve:
2x 2 − x
−x 2 − 2x = (2x − 1)x
4
2
-3
-2
-1
1
-2
-4
Solving Equations
Section A.6
2
3
Read section A.8 before next lecture.
Solving Equations
Section A.6