Exercise 23 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.

8x^2 + 12x = 8

SubEqn

LHS-8=RHS-8

8x^2 + 12x - 8 = 0

FactorOut

Factor out 4

4(2x^2 + 3x - 2) = 0

DivEqn

.LHS /4.=.RHS /4.

2x^2 + 3x - 2 = 0

Now, we can identify the values of a, b, and c. 2x^2 + 3x - 2 = 0 ⇕ 2x^2+ 3x+( - 2)=0 We see that a= 2, b= 3, and c= - 2. Let's substitute these values into the Quadratic Formula.

x=- b±sqrt(b^2-4ac)/2a

SubstituteValues

Substitute values

x=- 3±sqrt(3^2-4( 2)( - 2))/2( 2)

▼

Solve for x and Simplify

CalcPow

Calculate power

x=- 3±sqrt(9-4(2)(- 2))/2(2)

Multiply

Multiply

x=- 3±sqrt(9-8(- 2))/4

MultNegNegOnePar

- a(- b)=a* b

x=- 3±sqrt(9+16)/4

AddTerms

Add terms

x=- 3±sqrt(25)/4

CalcRoot

Calculate root

x=- 3 ± 5/4

The solutions for this equation are x= - 3± 54. Let's separate them into the positive and negative cases.

x=- 3 ± 5/4
x_1=- 3 + 5/4	x_2=- 3 - 5/4
x_1=2/4	x_2=- 8/4
x_1=0.5	x_2=- 2

Using the Quadratic Formula, we found that the solutions of the given equation are x_1=0.5 and x_2=- 2.