Exercise 27 Page 137

We will use the Quadratic Formula to solve the given quadratic equation. ax^2+ bx+ c=0 ⇕ x=- b± sqrt(b^2-4 a c)/2 a Let's start by rewriting the equation so all of the terms are on the left-hand side and then simplify as much as possible.

2x^2 = 12x - 18

SubEqn

LHS-12x=RHS-12x

2x^2 - 12x = - 18

AddEqn

LHS+18=RHS+18

2x^2 - 12x + 18 = 0

FactorOut

Factor out 2

2(x^2 - 6x + 9) = 0

DivEqn

.LHS /2.=.RHS /2.

x^2 - 6x + 9 = 0

Now, we can identify the values of a, b, and c. x^2 - 6x + 9 = 0 ⇕ 1x^2+( - 6)x+ 9=0 We see that a= 1, b= - 6, and c= 9. Let's substitute these values into the Quadratic Formula.

x=- b±sqrt(b^2-4ac)/2a

SubstituteValues

Substitute values

x=- ( -6)±sqrt(( - 6)^2-4( 1)( 9))/2( 1)

▼

Solve for x and Simplify

NegNeg

- (- a)=a

x=6±sqrt((- 6)^2-4(1)(9))/2(1)

CalcPow

Calculate power

x=6±sqrt(36-4(1)(9))/2(1)

Multiply

Multiply

x=6±sqrt(36-36)/2

SubTerm

Subtract term

x=6±sqrt(0)/2

CalcRoot

Calculate root

x=6± 0/2

Since adding or subtracting zero does not change the value of a number, the numerator will simplify to 6. Therefore, we will get only one value of x. x = 6/2 ⇔ x = 3 Using the Quadratic Formula, we found that the solution of the given equation is x = 3.