algebra 1 lesson 9-5 warm-up. algebra 1 “factoring to solve quadratic equations” (9-5) how do...
DESCRIPTION
“Factoring to Solve Quadratic Equations” (9-5) How do you solve a quadratic equation when b 0? Rule: To solve a quadratic equation (ax 2 + bx + c = 0) in which b 0, you can use the Zero Product property which says: If ab = 0, then a = 0 or b = 0 You can apply the Zero Product Rule to find solutions to a quadratic equation by factoring the equation, making one of the factors equal 0, and solving for the other factor. Example: Solve (x + 2)(x - 3) = 0 Method 1: Zero Product Rule: If (x + 2)(x - 3) = 0, then either (x + 2) = 0 or (x - 3) = 0 according to the Zero Product Rule. x + 2 = 0 Let (x + 2) = Subtract 2 from both sides x = -2 x - 3 = 0 Let (x - 3) = Subtract 2 from both sides x = 3 The solutions are x = -2 and 3.TRANSCRIPT
ALGEBRA 1
Lesson 9-5 Warm-Up
ALGEBRA 1
ALGEBRA 1
“Factoring to Solve Quadratic Equations” (9-5)
How do you solve a quadratic equation when b 0?
Rule: To solve a quadratic equation (ax2 + bx + c = 0) in which b 0, you can use the Zero Product property which says:
If ab = 0, then a = 0 or b = 0
You can apply the Zero Product Rule to find solutions to a quadratic equation by factoring the equation, making one of the factors equal 0, and solving for the other factor.Example: Solve (x + 2)(x - 3) = 0
Method 1: Zero Product Rule: If (x + 2)(x - 3) = 0, then either (x + 2) = 0 or (x - 3) = 0 according to the Zero Product Rule.
x + 2 = 0 Let (x + 2) = 0 -2 -2 Subtract 2 from both sidesx = -2x - 3 = 0 Let (x - 3) = 0 +3 +3 Subtract 2 from both sidesx = 3
The solutions are x = -2 and 3.
ALGEBRA 1
“Factoring to Sove Quadratic Equations” (9-5)
Method 2: Identity Property of Multiplication: : If (x + 2)(x - 3) = 0, and we divide both sides by one of the factors, (x + 2) or (x - 3), then the result is the other factor equals 0.
So, the solutions are -2 and 3.
x – 3 = 0 Identity Property 1(x – 3) = x – 3 + 3 +3 Add 3 to both sides. x = 3
x + 2 = 0 Identity Property 1(x + 2) = x + 2 - 2 -2 Subtract 2 from both sides. x = -2
ALGEBRA 1
“Factoring to Sove Quadratic Equations” (9-5)
How do you check solutions to a quadratic equations?
Check: To check the solutions of a quadratic equaltion, substitute the variable for each solution separately and see if you get 0 to make the equation a true statement..
(x + 2)(x - 3) = 0 Given (-2 + 2)(-2 - 3) = 0 Substitute x = -2 into the factored
equation (0)(-5) = 0 Simplify 0 = 0 True Statement
(x + 2)(x - 3) = 0 Given (3 + 2)(3 - 3) = 0 Substitute x =3 into the factored
equation (5)(0) = 0 Simplify 0 = 0 True Statement
ALGEBRA 1
Solve (2x + 3)(x – 4) = 0 by using the Zero-Product Property.
(2x + 3)(x – 4) = 0
2x + 3 = 0 or x – 4 = 0 Use the Zero-Product Property.
2x = –3 Solve for x.
Substitute 4 for x.
(2x + 3)(x – 4) = 0
[2(4) + 3][(4) – 4] 0
(11)(0) = 0
x = – 32 or x = 4
Check: Substitute – for x.32
(2x + 3)(x – 4) = 0
[2(– ) + 3][(– ) – 4] 032
32
(0)(– 5 ) = 012
Factoring to Solve Quadratic EquationsLESSON 9-5
Additional Examples
ALGEBRA 1
Solve x2 + x – 42 = 0 by factoring.
x2 + x – 42 = 0
(x + 7)(x – 6) = 0 Factor x2 + x – 42.
x + 7 = 0 or x – 6 = 0 Use the Zero-Product Property.
x = –7 or x = 6 Solve for x.
Factoring to Solve Quadratic EquationsLESSON 9-5
Additional Examples
ALGEBRA 1
“Factoring to Sove Quadratic Equations” (9-5)
How do you solve a quadratic equation not in standard form??
To solve a quadratic equation not in standard form, such as ax2 + bx = c, make the right side equal to zero by “undoing” everything from the right side. Once the equation is in standard form, factor it, and then solve for each factor as before.
ax2 + bx = c ax2 + bx = c Subtract c from both sides to undo it
-c -c from the right side. ax2 + bx – c = 0 Standard form.
Example: Solve 2x2 - 5x =88 2x2 - 5x = 88 Given
-88 -88 Subtract 88 from both sides 2x2 - 5x -88 = 0 Simplify (2x + 11)(x – 8) = 0 Factor 2x2 - 5x =88 2x + 11 = 0 x - 8 = 0 Let (2x + 11) = 0 and (x - 8) = 0 - 11 -11 +8 +8 Add 8 or subtract 11 from both sides 2x = -11 x = 8 2 2 Divide each side by 2. x = - or -5.511
2
ALGEBRA 1
Solve 3x2 – 2x = 21 by factoring.
3x2 – 2x – 21 = 0 Subtract 21 from each side.
(3x + 7)(x – 3) = 0 Factor 3x2 – 2x – 21.
3x + 7 = 0 or x – 3 = 0 Use the Zero-Product Property.
3x = –7 Solve for x.
x = – or x = 373
Factoring to Solve Quadratic EquationsLESSON 9-5
Additional Examples
ALGEBRA 1
The diagram shows a pattern for an open-top box. The total area of the sheet of materials used to make the box is 130 in.2. The height of the box is 1 in. Therefore, 1 in. 1 in. squares are cut from each corner. Find the dimensions of the box.
Define: Let x = width of a side of the box.Then the width of the material = x + 1 + 1 = x + 2The length of the material = x + 3 + 1 + 1 = x + 5
Words: length width = area of the sheet
Factoring to Solve Quadratic EquationsLESSON 9-5
Additional Examples
ALGEBRA 1
(continued)
Equation: (x + 2) (x + 5) = 130
x2 + 7x + 10 = 130 Find the product (x + 2) (x + 5).
x2 + 7x – 120 = 0 Subtract 130 from each side.
(x – 8) (x + 15) = 0 Factor x2 + 7x – 120.
x – 8 = 0 or x + 15 = 0 Use the Zero-Product Property.
x = 8 or x = –15 Solve for x.
The only reasonable solution is 8. So the dimensions of the box are 8 in. 11 in. 1 in.
Factoring to Solve Quadratic EquationsLESSON 9-5
Additional Examples
ALGEBRA 1
1. Solve (2x – 3)(x + 2) = 0.
Solve by factoring.
2. 6 = a2 – 5a 3. 12x + 4 = –9x2 4. 4y2 = 25
–2, 32
–1, 6 – 23 ± 5
2
Factoring to Solve Quadratic EquationsLESSON 9-5
Lesson Quiz