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McGraw Hill Integrated II, 2012

MH

McGraw Hill Integrated II, 2012 View details

8. Equations of Circles

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Exercise 52 Page 781

Make sure you rewrite the equation before factoring, leaving all the terms on one side.

H

Practice makes perfect

We want to solve the given equation. Let's use factoring to do so.

Factoring

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

n^2-4n=21

LHS-21=RHS-21

n^2-4n-21=0

Write as a difference

n^2+3n-7n-21=0

▼

Factor out n & - 7

Factor out n

n(n+3)-7n-21=0

Factor out - 7

n(n+3)-7(n+3)=0

Factor out (n+3)

(n+3)(n-7)=0

Solving

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

(n+3)(n-7)=0

Use the Zero Product Property

lcn+3=0 & (I) n-7=0 & (II)

(I): LHS-3=RHS-3

ln=- 3 n-7=0

(II): LHS+7=RHS+7

ln_1=- 3 n_2=7

The obtained solutions correspond to option H.

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 n=- 3.

n^2-4n=21

n= - 3

( -3)^2-4( -3)? =21

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Simplify

NegBaseToPosBase

(- a)^2=a^2

9-4(-3)? =21

MultNegNegOnePar

- a(- b)=a* b

9+12? =21

Add terms

21=21 ✓

Substituting and simplifying created a true statement, so we know that n=- 3 is a solution of the equation. Let's move on to n=7.

n^2-4n=21

n= 7

7^2-4( 7)? =21

▼

Simplify

Calculate power

49-4(7)? =21

Multiply

49-28? =21

Subtract term

21=21 ✓

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

Subchapter links

Exercises p.778-781