Exercise 2 Page 386

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. 0.75q^2=108 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.

0.75q^2=108

DivEqn

.LHS /0.75.=.RHS /0.75.

0.75q^2/0.75=108/0.75

FactorOut

Factor out 0.75

0.75q^2/0.75=0.75(144)/0.75

CancelCommonFac

Cancel out common factors

0.75q^2/0.75=0.75(144)/0.75

SimpQuot

Simplify quotient

q^2=144

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

q^2=144

SqrtEqn

sqrt(LHS)=sqrt(RHS)

sqrt(q^2)=sqrt(144)

sqrt(a^2)=± a

q=± sqrt(144)

CalcRoot

Calculate root

q=± 12

The solutions are q=-12 and q=12.