Exercise 32 Page 645

We are given three equations. A.& sqrt(4)=sqrt(x)+sqrt(2) B.& 4=x+2 C.& 2-sqrt(2)=sqrt(x) We will find which one of these equations has the same solution set as the radical equation sqrt(4)=sqrt(x+2). To do so, let's first solve the equation sqrt(4)=sqrt(x+2).

sqrt(4)=sqrt(x+2)

RaiseEqn

LHS^2=RHS^2

(sqrt(4))^2=(sqrt(x+2))^2

PowSqrt

( sqrt(a) )^2 = a

4=x+2

SubEqn

LHS-2=RHS-2

2=x

RearrangeEqn

Rearrange equation

x=2

The solution set of the equation consists of 2. We will now solve the equations given in options A, B, and C to find which one has the solution set consisting of 2. Let's first consider the equation in A.

sqrt(4)=sqrt(x)+sqrt(2)

SubEqn

LHS-sqrt(2)=RHS-sqrt(2)

sqrt(4)-sqrt(2)=sqrt(x)

CalcRoot

Calculate root

2-sqrt(2)=sqrt(x)

Notice that we obtained the equation in option C in the third step while solving the equation in option A. Therefore, their solution sets are the same.

2-sqrt(2)=sqrt(x)

▼

Solve for x

RaiseEqn

LHS^2=RHS^2

(2-sqrt(2))^2=(sqrt(x))^2

PowSqrt

( sqrt(a) )^2 = a

(2-sqrt(2))^2=x

ExpandNegPerfectSquare

(a-b)^2=a^2-2ab+b^2

4-4sqrt(2)+2=x

AddTerms

Add terms

6-4sqrt(2)=x

UseCalc

Use a calculator

0.343145...=x

RearrangeEqn

Rearrange equation

x=0.343145...

RoundDec

Round to 1 decimal place(s)

x≈ 0.3

As we can see, the solution to the equations in A and B is approximately 0.3, but not 2. It means that it is not the same as the solution of the equation sqrt(4)=sqrt(x+2). Let's now find the solution set of the equation in B.

4=x+2

SubEqn

LHS-2=RHS-2

2=x

RearrangeEqn

Rearrange equation

x=2

We can see that 2 is the solution of the equation 4=x+2, so this equation has the same solution set as the equation sqrt(4)=sqrt(x+2). Therefore, the correct choice is option B.