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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 28 Page 49

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

- 5, 18

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.

y^2-90=13y

LHS-13y=RHS-13y

y^2-13y-90=0

Write as a difference

y^2+5y-18y-90=0

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Factor out y & - 18

Factor out y

y(y+5)-18y-90=0

Factor out - 18

y(y+5)-18(y+5)=0

Factor out (y+5)

(y+5)(y-18)=0

Solving

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

(y+5)(y-18)=0

Use the Zero Product Property

lcy+5=0 & (I) y-18=0 & (II)

(I): LHS-5=RHS-5

ly=- 5 y-18=0

(II): LHS+18=RHS+18

ly_1=- 5 y_2=18

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^2-90=13y

y= - 5

( -5)^2-90? =13( - 5)

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Simplify

Calculate power

25-90? =13(- 5)

a(- b)=- a * b

25-90? =- 65

Subtract term

- 65=-65 ✓

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=18.

y^2-90=13y

y= 18

18^2-90? =13( 18)

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Simplify

Calculate power

324-90? =13(18)

Multiply

324-90? =234

Subtract term

234=234 ✓

Again, we created a true statement. y=18 is indeed a solution of the equation.

Subchapter links

Exercises p.49-51