Exercise 1 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. 7z^2=252 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.

7z^2=252

DivEqn

.LHS /7.=.RHS /7.

7z^2/7=252/7

FactorOut

Factor out 7

7z^2/7=7(36)/7

CancelCommonFac

Cancel out common factors

7z^2/7=7(36)/7

SimpQuot

Simplify quotient

z^2=36

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

z^2=36

SqrtEqn

sqrt(LHS)=sqrt(RHS)

sqrt(z^2)=sqrt(36)

sqrt(a^2)=± a

z=± sqrt(36)

CalcRoot

Calculate root

z=± 6

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