Exercise 24 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. - 42=7b^2-385 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.

- 42=7b^2-385

AddEqn

LHS+385=RHS+385

343=7b^2

DivEqn

.LHS /7.=.RHS /7.

343/7=7b^2/7

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Simplify quotient

FactorOut

Factor out 7

7(49)/7=7b^2/7

CancelCommonFac

Cancel out common factors

7(49)/7=7b^2/7

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.