Exercise 5 Page 380

To solve an equation, we should first gather all of the variable terms on one side of the equation and all of the constant terms on the other side, using the Properties of Equality. 3(2x+1)^2=27 In this case, we need to start by expanding the perfect square trinomial to simplify the left-hand side of the equation. Now we can continue to solve using the Properties of Equality.

12x^2 + 12x + 3 =27

SubEqn

LHS-3=RHS-3

12x^2 + 12x =24

DivEqn

.LHS /12.=.RHS /12.

x^2 + x =2

Now, we will solve the quadratic equation. x^2 + x =2 ⇔ 1x^2+( 1)x+( - 2)=0 We see above that a= 1, b= 1, and c= - 2. Let's substitute these values into the Quadratic Formula to solve the equation.

x=- b±sqrt(b^2-4ac)/2a

SubstituteValues

Substitute values

x=- 1±sqrt(1^2-4( 1)( - 2))/2( 1)

▼

Simplify right-hand side

BaseOne

1^a=1

x=- 1±sqrt(1 - 4(1)(- 2))/2(1)

Multiply

Multiply

x=- 1±sqrt(1+8)/2

AddTerms

Add terms

x=- 1±sqrt(9)/2

CalcRoot

Calculate root

x=- 1 ± 3/2

Now we can calculate the first root using the positive sign and the second root using the negative sign.

x=- 1 ± 3/2
x= - 1 + 3/2	x= - 1 -3/2
x= 2/2	x= - 4/2
x=1	x=- 2

The solution to the equation is x=- 2 or x= 1. We can write our answer in set-builder notation. { -2, 1 } This corresponds to answer B.