Exercise 29 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 - 8x = 12

SubEqn

LHS-12=RHS-12

2x^2 - 8x - 12 = 0

FactorOut

Factor out 2

2(x^2 - 4x - 6) = 0

DivEqn

.LHS /2.=.RHS /2.

x^2 - 4x - 6 = 0

Now, we can identify the values of a, b, and c. x^2 - 4x - 6 ⇕ 1x^2+( - 4)x+( - 6)=0 We see that a= 1, b= - 4, and c= - 6. Let's substitute these values into the Quadratic Formula.

x=- b±sqrt(b^2-4ac)/2a

SubstituteValues

Substitute values

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

▼

Solve for x and Simplify

NegNeg

- (- a)=a

x=4±sqrt((- 4)^2-4(1)(- 6))/2(1)

CalcPow

Calculate power

x=4±sqrt(16-4(1)(- 6))/2(1)

MultByOne

a * 1=a

x=4±sqrt(16-4(- 6))/2

MultNegNegOnePar

- a(- b)=a* b

x=4±sqrt(16 + 24)/2

AddTerms

Add terms

x=4±sqrt(40)/2

SplitIntoFactors

Split into factors

x=4±sqrt(4* 10)/2

SqrtProd

sqrt(a* b)=sqrt(a)*sqrt(b)

x=4± sqrt(4)* sqrt(10)/2

CalcRoot

Calculate root

x=4± 2 sqrt(10)/2

FactorOut

Factor out 2

x=2(2± sqrt(10))/2

CancelCommonFac

Cancel out common factors

x=2± sqrt(10)

The solutions for this equation are x=2 ± sqrt(10). Let's separate them into the positive and negative cases.

x=2 ± sqrt(10)
x_1=2 - sqrt(10)	x_2=2 + sqrt(10)
x_1 ≈ - 1.2	x_2 ≈ 5.2

Using the Quadratic Formula, we found that the solutions of the given equation are x_1 ≈ - 1.2 and x_2 ≈ 5.2.