Exercise 41 Page 568

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+x=12 ⇓ x^2+x-12=0 Now we can identify the values of a, b, and c. x^2+x-12=0 ⇕ 1x^2+ 1x+( - 12)=0 We see that a= 1, b= 1, and c= - 12. Let's substitute these values into the Quadratic Formula.

x=- b±sqrt(b^2-4ac)/2a

SubstituteValues

Substitute values

x=- 1±sqrt(1^2-4( 1)( - 12))/2( 1)

▼

Solve for x and Simplify

CalcPow

Calculate power

x=- 1±sqrt(1-4(1)(- 12))/2(1)

Multiply

Multiply

x=- 1±sqrt(1-4(- 12))/2

MultNegNegOnePar

- a(- b)=a* b

x=- 1±sqrt(1+48)/2

AddTerms

Add terms

x=- 1±sqrt(49)/2

CalcRoot

Calculate root

x=- 1± 7/2

The solutions for this equation are x= - 1± 72. Let's separate them into the positive and negative cases.

x=- 1± 7/2
x_1=- 1+7/2	x_2=- 1-7/2
x_1=6/2	x_2=- 8/2
x_1=3	x_2=- 4

Using the Quadratic Formula, we found that the solutions of the given equation are x_1=3 and x_2=- 4.