Exercise 46 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. 3h^2=h^2+18 First, we need to isolate the variable on one side of the equation. To do so, we will use the Subtraction Property of Equality and Division Property of Equality.

3h^2=h^2+18

SubEqn

LHS-h^2=RHS-h^2

3h^2-h^2=h^2+18-h^2

SubTerm

Subtract term

2h^2=18

DivEqn

.LHS /2.=.RHS /2.

2h^2/2=18/2

▼

Simplify quotient

FactorOut

Factor out 2

2h^2/2=2(9)/2

CancelCommonFac

Cancel out common factors

2h^2/2=2(9)/2

SimpQuot

Simplify quotient

h^2/1=9/1

DivByOne

a/1=a

h^2=9

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

h^2=9

SqrtEqn

sqrt(LHS)=sqrt(RHS)

sqrt(h^2)=sqrt(9)

sqrt(a^2)=± a

h=± sqrt(9)

CalcRoot

Calculate root

h=± 3

The solutions are h=- 3 and h=3.