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 21 Page 49

Make sure you rewrite the equation leaving all the terms on one side, and that you factor out the greatest common factor if it exists.

- 5, 4

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. Notice that we do not need to factor out the greatest common factor.

y^2+y=20

LHS-20=RHS-20

y^2+y-20=0

Write as a difference

y^2+5y-4y-20=0

▼

Factor out y & - 4

Factor out y

y(y+5)-4y-20=0

Factor out - 4

y(y+5)-4(y+5)=0

Factor out (y+5)

(y+5)(y-4)=0

Solving

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

(y+5)(y-4)=0

Use the Zero Product Property

lcy+5=0 & (I) y-4=0 & (II)

(I): LHS-5=RHS-5

ly=- 5 y-4=0

(II): LHS+4=RHS+4

ly_1=- 5 y_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 y=- 5.

y^2+y=20

y= - 5

( -5)^2+( -5)? =20

▼

Simplify

Calculate power

25+(-5)? =20

a+(- b)=a-b

25-5? =20

Subtract term

20=20 ✓

Substituting and simplifying created a true statement, so we know that y=- 5 is a solution of the equation. Let's move on to y=4.

y^2+y=20

y= 4

4^2+ 4? =20

▼

Simplify

Calculate power

16+4? =20

Add terms

20=20 ✓

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

Subchapter links

Exercises p.49-51