MTH 10905 Algebra

Solving Quadratic Equation using Factoring Chapter 5 Section 6

Recognize Quadratic Equation

 Quadratic Equation in Standard Form ax 2 + bx + c = 0 a, b, and c are real numbers a ≠ 0  Examples of a quadratic equation in standard form: x 2 + 4x – 12 = 0 2x 2 – 5x = 0 same as 2x 2 – 5x + 0 = 0 3x 2 – 2 = 0 same as 3x 2 + 0x – 2 = 0

Zero Factor Property

 ab = c What do you know about a and b if their product is 6?

 ab = 6 a = 6

b

and b = 6

a

What do I know about a and b if their product is 0? ab = 0 a = 0

b

a = 0 and b = 0

a

b = 0

Zero Factor Property

 Zero Factor Property a • b = 0, then a = 0 or b = 0  To solve a Quadratic Equation by factoring we use the Zero Factor Property.

 We know that is we multiply by 0 the product will be 0.  If a product equals 0 then one of the factors has to be 0.

Solve the Equation

Example: (x + 8) (x + 2) = 0 x + 8 = 0 and x + 2 = 0 x = -8 x = -2 To write (x+8)(x+2) in standard form you must multiply using FOIL. x 2 +10x +16 = 0 Does the above solutions solve the equation? YES (-8) 2 +10(-8) +16 = 0 (-2) 2 +10(-2) +16 = 0 64 – 80 + 16 = 0 4 – 20 + 16 = 0 -16 + 16 = 0 -16 + 16 = 0

Solve the Equation

Example: (2x + 5) (3x – 7) = 0 2x + 5 = 0 and 3x – 7 = 0 2x = -5 3x = 7

x

  5 2

x

 7 3 Check 1 st solution: (2x + 5) (3x – 7) = 0 [(2)(-5/2) + 5] [(3)(-5/2) – 7] = 0 [-10/2 + 5] [-7.5 – 7] = 0 [-5 + 5] [-14.5] = 0 [0] [-14.5] = 0 0 = 0

Solve the Equation

Example: (2x + 5) (3x – 7) = 0 2x + 5 = 0 and 3x – 7 = 0 2x = -5 3x = 7

x

  5 2

x

 7 3 Check 2 nd solution: (2x + 5) (3x – 7) = 0 [(2)(7/3) + 5] [(3)(7/3) – 7] = 0 [14/3 + 5] [21/3 – 7] = 0 [29/3] [7 – 7] = 0 [29/3] [0] = 0 0 = 0

Solving Quadratic Equation using Factoring

1.

Write the equation in standard form with the squared term having a positive coefficient. This will result in one side of the equation being 0.

2.

Factor the non-zero side.

3.

Set each factor containing a variable equal to zero (0) and solve the equation.

4.

Check each solution in the original equation.

Solve the Equation

Example: Check: 9x 9x 2 2 = 27x Put in standard form – 27x = 0 GCF = 9x 9x(x – 3) = 0 9x = 0 9x/9 = 0/9 x = 0 and x – 3 = 0 x-3+3 = 0+3 x = 3 9x 9(0) 2 2 = 27x = 27(0) 9x 9(3) 2 2 = 27x = 27(3) 0 = 0 9(9) = 81 81 = 81

Solve the Equation

Example: x x 2 2 + 11x + 34 = 4 + 11x + 30 = 0 (x + 5)(x + 6) = 0 x + 5 = 0 Put in standard form Factors 30 add 11 (5)(6) 6 + 5 and x + 6 = 0 x = 5 – 5 = 0 – 5 x = -5 x + 6 – 6 = 0 - 6 x = -6 Check: (-5) 2 x 2 + 11x + 34 = 4 + 11(-5) + 34 = 4 25 – 55 + 34 = 4 -30 + 34 = 4 4 = 4 (-6) 2 x 2 + 11x + 34 = 4 + 11(-6) + 34 = 4 36 – 66 + 34 = 4 -30 + 34 = 4

Solve the Equation

Example: Example: 5z 2 5z 2 5z 2 5(z 2 + 10z – 60 = -10z + 10z +10z – 60 = 0 + 20z – 60 = 0 + 4z – 12) = 0 5(x + 6)(x – 2) x + 6 = 0 Put in standard form GCF = 5 Factors of -12 add 4 (-2)(6) 6 + -2 and x – 2 = 0 x + 6 – 6 = 0 – 6 and x – 2 + 2 = 0 + 2 x = -6 x = 2 -x 2 -1(x + x + 6 = 0 2 – x – 6) = 0 -1(x + 2)(x – 3) = 0 Factor out -1 to make squared term positive Factors of -6 add -1 (2)(-3) 2 + -3 x + 2 = 0 x + 2 – 2 = 0 – 2 x = -2 and x – 3 = 0 and x – 3 + 3 = 0 + 3 x = 3 REMEMBER: We can check by putting the solutions back into the original equation.

Solve the Equation

Example: x 2 x 2 = 81 – 81 = 0 (x) 2 – (9) 2 = 0 (x + 9) (x – 9) = 0 x + 9 = 0 and x – 9 = 0 x = -9 x = 9 Put in standard form Difference in Two Squares a 2 – b 2 = (a + b) (a – b) Example: (x – 5)(x + 2) = 8 x 2 + 2x – 5x – 10 = 8 x 2 x x 2 2 – 3x – 10 = 8 – 3x – 10 – 8 = 0 – 3x – 18 = 0 (x + 3)(x – 6) = 0 x + 3 = 0 and x – 6 = 0 x = -3 x = 6 Multiply first, then put in standard form Factors of -18 add -3 (-6)(3) -6 + 3 REMEMBER: We can check by putting the solutions back into the original equation.

REMEMBER

 Always put the equation in standard form. ax 2 + bx + c = 0  The expression that is to be factored must be equal to 0 before the zero factor property can be used.

 Check your answer to the original equation.

HOMEWORK 5.6

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