Exercise 44 Page 379

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. First, we need to isolate the variable term on one side of the equation. To do so, we will first evaluate the power using inverse operations.

sqrt(36/9)=1/3m^2-10

CalcQuot

Calculate quotient

sqrt(4)=1/3m^2-10

CalcRoot

Calculate root

2=1/3m^2-10

Now, we can use the Addition Property of Equality and Multiplication Property of Equality to simplify the equation even further.

2=1/3m^2-10

AddEqn

LHS+10=RHS+10

12=1/3m^2

MultEqn

LHS * 3=RHS* 3

12(3)=1/3m^2(3)

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Simplify

Multiply

Multiply

36=1/3m^2(3)

CommutativePropMult

Commutative Property of Multiplication

36=3(1/3)m^2

a * 1/a=1

36=1m^2

IdPropMult

Identity Property of Multiplication

36=m^2

RearrangeEqn

Rearrange equation

m^2=36

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

m^2=36

SqrtEqn

sqrt(LHS)=sqrt(RHS)

sqrt(m^2)=sqrt(36)

sqrt(a^2)=± a

m=± sqrt(36)

CalcRoot

Calculate root

m=± 6

The solutions are m=- 6 and m=6.