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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 53 Page 230

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.

- 1, 103

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.

7x-3x^2=- 10

LHS+10=RHS+10

7x-3x^2+10=0

CommutativePropAdd

Commutative Property of Addition

- 3x^2+7x+10=0

Write as a sum

- 3x^2-3x+10x+10=0

▼

Factor out (- 3x) & 10

Factor out (- 3x)

- 3x(x+1)+10x+10=0

Factor out 10

- 3x(x+1)+10(x+1)=0

Factor out (x+1)

(x+1)(- 3x+10)=0

Solving

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

(x+1)(- 3x+10)=0

Use the Zero Product Property

lcx+1=0 & (I) - 3x+10=0 & (II)

(I): LHS-1=RHS-1

lx=- 1 - 3x+10=0

(II): LHS-10=RHS-10

lx=- 1 - 3x=- 10

(II): .LHS /(- 3).=.RHS /(- 3).

lx_1=- 1 x_2= 103

Checking Our Answer

Checking our answer

We can substitute our solutions back into the given equation and simplify to check if our answers are correct. We will start with x=- 1.

7x-3x^2=- 10

x= - 1

7( - 1)-3( - 1)^2? =- 10

▼

Evaluate left-hand side

Calculate power

7(- 1)-3(1)? =- 10

a(- b)=- a * b

- 7-3(1)? =- 10

Identity Property of Multiplication

- 7-3? =- 10

Subtract term

- 10=- 10 ✓

Substituting and simplifying created a true statement, so we know that x=- 1 is a solution of the equation. Let's move on to x= 103.

7x-3x^2=- 10

x= 10/3

7( 10/3)-3( 10/3)^2? =- 10

▼

Evaluate left-hand side

(a/b)^m=a^m/b^m

7(10/3)-3(100/9)? =- 10

MoveLeftFacToNum

a*b/c= a* b/c

70/3-300/9? =- 10

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

70/3-100/3? =- 10

Subtract fractions

-30/3? =- 10

Calculate quotient

- 10=- 10 ✓

Again, we created a true statement. x= 103 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