body{margin:0;} noscript{z-index: 10000; position: fixed; display: block; height: 100%; width: 100%; background: #fff;} noscript .fa{font-size: 40px; display:block; color: #daa520;} noscript div{margin-top: 6%; padding-left: 20px; padding-right: 20px; text-align: center;} ! You must have JavaScript enabled to use this site.

McGraw Hill Integrated II, 2012

MH

McGraw Hill Integrated II, 2012 View details

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

Continue to next subchapter

Exercise 20 Page 49

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

3, 4

Practice makes perfect

We want to solve the given equation by factoring.

Factoring

Let's start factoring the expression!

x^2-7x+12=0

Write as a difference

x^2-3x-4x+12=0

▼

Factor out x & - 4

Factor out x

x(x-3)-4x+12=0

Factor out - 4

x(x-3)-4(x-3)=0

Factor out (x-3)

(x-3)(x-4)=0

Solving

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

(x-3)(x-4)=0

Use the Zero Product Property

lcx-3=0 & (I) x-4=0 & (II)

(I): LHS+3=RHS+3

lx=3 x-4=0

(II): LHS+4=RHS+4

lx_1=3 x_2=4

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=3.

x^2-7x+12=0

x= 3

3^2-7( 3)+12? =0

▼

Simplify

Calculate power

9-7(3)+12? =0

Multiply

9-21+12? =0

Add and subtract terms

0=0 ✓

Substituting and simplifying created a true statement, so we know that x=3 is a solution of the equation. Let's move on to x=4.

x^2-7x+12=0

x= 4

4^2-7( 4)+12? =0

▼

Simplify

Calculate power

16-7(4)+12? =0

Multiply

16-28+12? =0

Add and subtract terms

0=0 ✓

Again, we created a true statement. x=4 is indeed a solution of the equation.

Subchapter links

Exercises p.49-51