Bellwork #34 April 24th, 2017 Grab a BW sheet from the Algebra I bin. Quiz on Sections 10.1 – 10.5 on Thursday! A review will be given tomorrow.

Section 10.4 Review Solving Quadratics (when b=0): Step 1: Isolate the term that is being square on one side of the equation. Step 2: Square root both sides of the equation. Step 3: Finish solving for x, if necessary.

Find the EXACT solutions to the following equations: 1.)

2.) (x - 6)2 - 14 = 3

3.) 2(x + 1)2 + 3 = 51

Standard Form of a Quadratic Function: y = ax2 + bx + c Standard Form of a Quadratic Equation: 0 = ax2 + bx + c

10.5 – Factoring to Solve Quadratic Equations Algebra I

Objective 1: Solving Quadratic Equations by Factoring In the previous lesson, you solved quadratic equations by finding square roots. This only works when b = 0. You can solve quadratic equations by using the Zero Product Property, if b ≠ 0. Key Concepts: Property – Zero-Product Property: For every real number a and b, if ab = 0, then a = 0 or b = 0. Example: If (x+3)(x+2) = 0, then x + 3 = 0 or x + 2 =0

Example 1: Using the Zero-Product Property Solve each equation: 1.) (x + 3)(x – 8) = 0 x =

Quick Check #1: 2.) (4x + 7)(x – 6) = 0 3.) (3x – 1)(x + 11) = 0 x = x =

Quick Check #2: 4.) 2x(x – 2) = 0 x =

Answer: Factoring Quadratics!! You can also use the Zero-Product Property to solve equations of the form 𝑎 𝑥 2 +𝑏𝑥+𝑐=0 if the quadratic expression 𝑎 𝑥 2 +𝑏𝑥+𝑐 can be factored.

Example 2: Solve for x by factoring 5.) 4 𝑥 2 −12𝑥−27

Quick Check #3: Solve for x by factoring 6.) 𝑥 2 −8=2𝑥

7.) 7 𝑥 2 −21𝑥=0

8.) 6 𝑥 2 +5𝑥−21=0

9.) 3 𝑥 2 −11=2 𝑥 2 −4𝑥+1

10.) 2 𝑥 3 −5 𝑥 2 =18𝑥−45

HW #24: due tomorrow! Section 10.4 – pg. 531 #s 1-2, 5, 10-18 Section 10.5 – pgs. 538 – 539 #s 1-3, 9, 10, 13, 14, 17, 27, 32