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Pearson Algebra 2 Common Core, 2011

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Pearson Algebra 2 Common Core, 2011 View details

5. Quadratic Equations

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Exercise 14 Page 229

Make sure you rewrite the equation leaving all the terms on one side and that you factor out the greatest common factor if it exists.

- 23, 6

Practice makes perfect

We want to solve the given equation by factoring.

Factoring

Let's start by writing all the terms on one side of the equals sign.

3x^2=16x+12

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Simplify

LHS-16x=RHS-16x

3x^2-16x=12

LHS-12=RHS-12

3x^2-16x-12=0

Write as a sum

3x^2-18x+2x-12=0

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Factor out 3x & 2

Factor out 3x

3x(x-6)+2x-12=0

Factor out 2

3x(x-6)+2(x-6)=0

Factor out (x-6)

(3x+2)(x-6)=0

Solving

To solve this equation, we will apply the Zero Product Property.

(3x+2)(x-6)=0

Use the Zero Product Property

lc3x+2=0 & (I) x-6=0 & (II)

(I): LHS-2=RHS-2

l3x=- 2 x-6=0

(I): .LHS /3.=.RHS /3.

lx=- 23 x-6=0

(II): LHS+6=RHS+6

lx_1=- 23 x_2=6

Checking

We can substitute our solutions back into the given equation and simplify to check if our answers are correct. We will start with x=- 23.

3x^2=16x+12

x= -2/3

3( -2/3)^2? =16( -2/3)+12

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Simplify

Calculate power

3(4/9)? =16(-2/3)+12

MoveRightFacToNum

a/c* b = a* b/c

3* 4/9? =16(-2/3)+12

a/b=.a /3./.b /3.

4/3? =16(-2/3)+12

a(- b)=- a * b

4/3? =-32/3+12

a = 3* a/3

4/3? =-32/3+36/3

Add fractions

4/3=4/3 ✓

Substituting and simplifying created a true statement, so we know that x=- 23 is a solution of the equation. Let's move on to x=6.

3x^2=16x+12

x= 6

3( 6)^2? =16( 6)+12

▼

Simplify

Calculate power

3(36)? =16(6)+12

Multiply

108? =96+12

Add terms

108=108 ✓

Again, we created a true statement. x=6 is indeed a solution of the equation.

Subchapter links

Lesson Check p.229

Practice and Problem-Solving Exercises p.229-231

Standardized Test Prep p.231

Mixed Review p.231