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**Solving Quadratic Equations**

Zero Product Property Using Factoring to solve Quadratic Equations

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Degree of an equation Equations with one variable, the degree is equal to the highest exponent 3x + 4 = first degree equation second degree equation, also called quadratic equations

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**Solving a First Degree Equation**

You have had a good amount of experience with this: 3x + 4 = to solve, get the variable alone subtract 4 on both sides 3x = divide both sides by 3 x = variable alone, coefficient of 1 3(3) + 4 = is the only number that makes = this equation true

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**Solving a Second Degree Equation**

Getting the variable alone, with a coefficient of one will work in some 2nd degree equations. subtract 2 on both sides divide both sides by 4 take the square root of both sides 2 or -2 can make the equation true

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**Solving a 2nd Degree Equation**

What about this one? Or this one? We need new strategy…

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Lets go over some vocabulary 2nd degree equations—we are going to call them quadratic equations or quadratics

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Lets go over some vocabulary Factoring a number or expression— means to break it down into two or more parts that are multiplied together.

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Zero Product Property If A • 5 = 0 then A = 0

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Zero Product Property If • B = 0 then B = 0

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**Zero Product Property If A • B = 0 then A = 0 or B = 0,**

0 • B = or A • 0 = 0 or both A and B equal 0 0 • 0 = 0

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**Solve (x + 3)(x - 5) = 0 If A • B = 0**

We can use the Zero Product Property whenever we have two factors that equal zero we know that either A = or B = 0 x + 3 = or x - 5 = 0 Solve each equation. x = or x = 5

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**Solve (2a + 4)(a + 7) = 0 A • B = 0 So…. 2a + 4 = 0 or a + 7 = 0**

{-2, -7}

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**A(B) = 0 So…….. t = 0 or t - 3 = 0 or t = 3 {0, 3}**

Solve t(t - 3) = 0 A(B) = 0 So…….. t = or t - 3 = 0 or t = 3 {0, 3}

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Solve (y – 3)(2y + 6) = 0 {-3, 3} {-3, 6} {3, 6} {3, -6}

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**Solving a Quadratic Equation**

What about this one? Or this one? Lets take them one at a time

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**Solving a Quadratic Equation**

What about this one? if we are going to use the zero product property, it needs to equal zero It also has to be the product of 2 factors We have to factor it Find GCF, then divide by it

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**Solving a Quadratic Equation**

What about this one? if we are going to use the zero product property, it needs to equal zero Now break up the two factors and make each equal to zero Then solve each

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**Solve x2 - 11x = 0 GCF = x x(x - 11) = 0 x = 0 or x - 11 = 0**

{0, 11}

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**Solve a2 - 24a +144 = 0 a2 - 24a + 144 = 0 (a - 12)(a - 12) = 0**

This one does not have a GCF other than 1. We use the X box to factor and solve this one. a2 - 24a = 0 (a - 12)(a - 12) = 0 a - 12 = 0 a = 12 {12} Put in descending order Squared term has to be positive Put 1st and 3rd term in the box We have two boxes and 1 term left, the xgame will show us how to split them up. Multiply the 1st and 3rd term and put on top of x The middle term goes on the bottom The numbers you find go in the two remaining boxes Find the GCF of each column and row for your factors a -12 144a2 a -12a a2 -12a -12a -12 -12a 144 -24a

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Solve x2 + 2x = 15 This one does not have a GCF other than 1. We use the X box to factor and solve this one. x2 + 2x – 15 = 0 x2 + 2x – 15 = 0 (x - 3)(x + 5) = 0 x – 3 = x + 5 = 0 x = 3 or x = -5 {3, -5} Put in descending order Squared term has to be positive Put 1st and 3rd term in the box We have two boxes and 1 term left, the xgame will show us how to split them up. Multiply the 1st and 3rd term and put on top of x The middle term goes on the bottom The numbers you find go in the two remaining boxes Find the GCF of each column and row for your factors x +5 -15x2 x x2 5x 5x -3x -3 -3x -15 2x

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**4 steps for solving a quadratic equation by factoring**

Set the equation equal to 0. 2 terms, factor by distribution or or difference of two squares 4 terms, reverse box 3 terms, X box No matter how many terms, always start by finding the GCF

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