Exercise 23 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. d^2+5=41 First, we need to isolate the variable term on one side of the equation. To do so, we will use the Subtraction Property of Equality.

d^2+5=41

SubEqn

LHS-5=RHS-5

d^2+5-5=41-5

SubTerm

Subtract term

d^2=36

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

d^2=36

SqrtEqn

sqrt(LHS)=sqrt(RHS)

sqrt(d^2)=sqrt(36)

sqrt(a^2)=± a

d=± sqrt(36)

CalcRoot

Calculate root

d=± 6

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