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McGraw Hill Integrated II, 2012

MH

McGraw Hill Integrated II, 2012 View details

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

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Exercise 7 Page 49

Start by factoring the equation. Then, use the Zero Product Property to solve it.

6, 9

Practice makes perfect

We want to solve the given equation by factoring.

Factoring

Let's start factoring the expression!

x^2-15x+54=0

Write as a difference

x^2-6x-9x+54=0

▼

Factor out x & - 9

Factor out x

x(x-6)-9x+54=0

Factor out - 9

x(x-6)-9(x-6)=0

Factor out (x-6)

(x-6)(x-9)=0

Solving

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

(x-6)(x-9)=0

Use the Zero Product Property

lcx-6=0 & (I) x-9=0 & (II)

(I): LHS+6=RHS+6

lx=6 x-9=0

(II): LHS+9=RHS+9

lx_1=6 x_2=9

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=6.

x^2-15x+54=0

x= 6

6^2-15( 6)+54? =0

▼

Simplify

Calculate power

36-15(6)+54? =0

Multiply

36-90+54? =0

Add and subtract terms

0=0 ✓

Substituting and simplifying created a true statement, so we know that x=6 is a solution of the equation. Let's move on to x=9.

x^2-15x+54=0

x= 9

9^2-15( 9)+54? =0

▼

Simplify

Calculate power

81-15(9)+54? =0

Multiply

81-135+54? =0

Add and subtract terms

0=0 ✓

Again, we created a true statement. x=9 is indeed a solution of the equation.

Subchapter links

Exercises p.49-51