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Pearson Algebra 1 Common Core, 2015

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Pearson Algebra 1 Common Core, 2015 View details

4. Factoring to Solve Quadratic Equations

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Exercise 1 Page 570

Use the Zero Product Property to solve the given equation.

4, 7

Practice makes perfect

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

(v-4)(v-7)=0

Use the Zero Product Property

lcv-4=0 & (I) v-7=0 & (II)

(I): LHS+4=RHS+4

lv=4 v-7=0

(II): LHS+7=RHS+7

lv_1=4 v_2=7

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 v=4.

(v-4)(v-7)=0

v= 4

( 4-4)( 4-7)? =0

▼

Simplify

Subtract terms

0(-3)? =0

Zero Property of Multiplication

0=0 ✓

Substituting and simplifying created a true statement, so we know that v=4 is a solution of the equation. Let's move on to v=7.

(v-4)(v-7)=0

v= 7

( 7-4)( 7-7)? =0

▼

Simplify

Subtract terms

3(0)? =0

Zero Property of Multiplication

0=0 ✓

Again, we created a true statement. v=7 is indeed a solution of the equation.

Subchapter links

Lesson Check p.570

Practice and Problem-Solving Exercises p.571-572