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McGraw Hill Integrated II, 2012

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McGraw Hill Integrated II, 2012 View details

6. Solving x^2+bx+c=0

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Exercise 29 Page 49

Make sure you rewrite the equation leaving all the terms on one side. Then, solve by factoring.

4, 12

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.

h^2+48=16h

LHS-16h=RHS-16h

h^2-16h+48=0

Write as a difference

h^2-4h-12h+48=0

▼

Factor out h & - 12

Factor out h

h(h-4)-12h+48=0

Factor out - 12

h(h-4)-12(h-4)=0

Factor out (h-4)

(h-4)(h-12)=0

Solving

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

(h-4)(h-12)=0

Use the Zero Product Property

lch-4=0 & (I) h-12=0 & (II)

(I): LHS+4=RHS+4

lh=4 h-12=0

(II): LHS+12=RHS+12

lh_1=4 h_2=12

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 h=4.

h^2+48=16h

h= 4

4^2+48? =16( 4)

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Simplify

Calculate power

16+48? =16(4)

Multiply

16+48? =64

Add terms

64=64 ✓

Substituting and simplifying created a true statement, so we know that h=4 is a solution of the equation. Let's move on to h=12.

h^2+48=16h

h= 12

12^2+48? =16( 12)

▼

Simplify

Calculate power

144+48? =16(12)

Multiply

144+48? =192

Add terms

192=192 ✓

Again, we created a true statement. h=12 is indeed a solution of the equation.

Subchapter links

Exercises p.49-51