Section A.6 Solving Equations Math 1051 - Precalculus I
A.6 Solving Equations Simplify 1 2x 6 x + 2 x 2 9
A.6 Solving Equations Simplify 1 2x 6 x + 2 x 2 9 Factor: 2x 6 = 2(x 3) x 2 9 = (x 3)(x + 3) LCD = 2(x 3)(x + 3)
A.6 Solving Equations Simplify 1 2x 6 x + 2 x 2 9 LCD = 2(x 3)(x + 3), so 1 2x 6 x + 2 x 2 9 = 1 2(x 3) x + 3 x + 3 x + 2 (x 3)(x + 3) 2 2 (x + 3) (2x + 4) = 2(x 3)(x + 3) x 1 = 2(x 3)(x + 3)
Solving Equations Equation: Two expressions set equal to each other.
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 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.
Three types of solutions
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.
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.
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.
2 3 (2x + 3) = 1 (3 x) 1 2
2 3 (2x + 3) = 1 (3 x) 1 2 5 6 4 2 2 4 6 5
5(x 4) (3 x) = 2(x + 5) + 4x
5(x 4) (3 x) = 2(x + 5) + 4x 4 2 6 4 2 2 4 6 2 4
5(x 4) (3 x) = 2(x + 5) + 4x 4 2 6 4 2 2 4 6 2 4 No solution!
6x 2 5 = 13x
6x 2 5 = 13x 30 20 10 4 2 2 4
6x 2 5 = 13x 30 20 10 4 2 2 4 We can check that both of these work
Factoring 3x 2 + 30x = 6x 3
Factoring 3x 2 + 30x = 6x 3 100 50 4 2 2 4 50
Absolute Values 4 3x 4 = 20
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.
Absolute Values 4 3x 4 = 20 25 20 15 10 5
5 x 2 + 3x 2 3 = 7
5 x 2 + 3x 2 3 = 7 15 10 5 4 2 2 4
Square Roots (x 2) 2 = 9
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.
Square Roots (x 2) 2 = 9 8 6 4 2 6 4 2 2 4 6 2
Completing the Square x 2 + 6x = 4
Completing the Square x 2 + 6x = 4 Completing the square Since (x + a) 2 = x 2 + 2ax + a 2, let b = 2a so that b 2 quadratic equation like = a. In a x 2 + bx = c we can solve by adding ( b 2) 2 to both sides.
Completing the Square 2x 2 3x = 1
Completing the Square 2x 2 3x = 1 8 6 4 2 4 2 2 4 2
Quadratic Equation The standard form for a quadratic equation is ax 2 + bx + c = 0 where a, b, c are real numbers and a 0. We can use completing the square to solve this equation to get the quadratic formula: x 1,2 = b ± b 2 4ac 2a
Quadratic Equation 2x 2 3x = 1
Quadratic Equation 2x 2 3x = 1 Same answer as we found completing the square
How about the cubic equation ax 3 + bx 2 + cx + d = 0? The solution is:
x 1 = b 3a 1 3 1 3a 2 1 3 1 3a 2 [ ] 2b 3 9abc + 27a 2 d + (2b 3 9abc + 27a 2 d) 2 4(b 2 3ac) 3 [ ] 2b 3 9abc + 27a 2 d (2b 3 9abc + 27a 2 d) 2 4(b 2 3ac) 3 x 2 = b 3a 1 + i 3 3 1 + 6a 2 1 i 3 3 1 + 6a 2 [ ] 2b 3 9abc + 27a 2 d + (2b 3 9abc + 27a 2 d) 2 4(b 2 3ac) 3 [ ] 2b 3 9abc + 27a 2 d (2b 3 9abc + 27a 2 d) 2 4(b 2 3ac) 3 x 3 = b 3a 1 i 3 3 1 + 6a 2 1 + i 3 3 1 + 6a 2 [ ] 2b 3 9abc + 27a 2 d + (2b 3 9abc + 27a 2 d) 2 4(b 2 3ac) 3 [ ] 2b 3 9abc + 27a 2 d (2b 3 9abc + 27a 2 d) 2 4(b 2 3ac) 3
Gerolamo Cardano
Cubic Equations No magical formula. Hopefully we can factor. x 3 1 = 0
Cubic Equations No magical formula. Hopefully we can factor. x 3 1 = 0 5 4 2 2 4 5
Solve: ( ) ( ) 2x 2 x x 2 2x = (2x 1)x
Solve: ( ) ( ) 2x 2 x x 2 2x = (2x 1)x 4 2 3 2 1 1 2 3 2 4
Read section A.8 before next lecture.