Exercise 22 Page 376

Let's start by recalling how to solve equations that contain a variable that is squared and equal to a non-negative number. These types of equations have two solutions. x^2=a ⇒ x=± sqrt(a) This is because both (sqrt(a))^2 and (- sqrt(a))^2 are equal to a. Let's think of a more concrete example. sqrt(9)=±3 because 3^2=9 and (-3)^2=9 With this in mind, let's solve the given equation. 2r^2=162 First, we need to isolate the variable term on one side of the equation. To do so, we will use the Division Property of Equality.

2r^2=162

DivEqn

.LHS /2.=.RHS /2.

2r^2/2=162/2

FactorOut

Factor out 2

2r^2/2=2(81)/2

CancelCommonFac

Cancel out common factors

2r^2/2=2(81)/2

SimpQuot

Simplify quotient

r^2/1=81/1

DivByOne

a/1=a

r^2=81

Next, since r is raised to the second power, we will take the square root of both sides. Let's do it!

r^2=81

SqrtEqn

sqrt(LHS)=sqrt(RHS)

sqrt(r^2)=sqrt(81)

sqrt(a^2)=± a

r=± sqrt(81)

CalcRoot

Calculate root

r=± 9

The solutions are r=- 9 and r=9.