Exercise 15 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. 190=4b^2-6 First, we need to isolate the variable on one side of the equation. To do so, we will use the Addition Property of Equality and Division Property of Equality.

190=4b^2-6

AddEqn

LHS+6=RHS+6

196=4b^2

DivEqn

.LHS /4.=.RHS /4.

196/4=4b^2/4

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Simplify quotient

FactorOut

Factor out 4

4(49)/4=4b^2/4

CancelCommonFac

Cancel out common factors

4(49)/4=4b^2/4

SimpQuot

Simplify quotient

49/1=b^2/1

DivByOne

a/1=a

49=b^2

RearrangeEqn

Rearrange equation

b^2=49

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

b^2=49

SqrtEqn

sqrt(LHS)=sqrt(RHS)

sqrt(b^2)=sqrt(49)

sqrt(a^2)=± a

b=± sqrt(49)

CalcRoot

Calculate root

b=± 7

The solutions are b=- 7 and b=7.