Exercise 4 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. x^2 + 3x = 12 ⇔ x^2 + 3x - 12 = 0 Now, we can identify the values of a, b, and c. x^2 + 3x - 12 = 0 ⇕ 1x^2+ 3x+( - 12)=0 We see that a= 1, b= 3, and c= - 12. 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( 1)( - 12))/2( 1)

▼

Solve for x and Simplify

CalcPow

Calculate power

x=- 3±sqrt(9-4(1)(- 12))/2(1)

MultByOne

a * 1=a

x=- 3±sqrt(9-4(- 12))/2

MultNegNegOnePar

- a(- b)=a* b

x=- 3±sqrt(9 + 48)/2

AddTerms

Add terms

x=- 3±sqrt(57)/2

The solutions for this equation are x= - 3 ± sqrt(57)2. Let's separate them into the positive and negative cases.

x=- 3± sqrt(57)/2
x_1=- 3 + sqrt(57)/2	x_2=- 3 - sqrt(57)/2
x_1≈2.3	x_2≈- 7.3

Using the Quadratic Formula, we found that the solutions of the given equation are x_1≈ 2.3 and x_2 ≈ - 7.3.