Exercise 11 Page 713

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.

x^2=x+12

SubEqn

LHS-x=RHS-x

x^2-x=12

SubEqn

LHS-12=RHS-12

x^2-x-12=0

Now, we can identify the values of a, b, and c. x^2-x-12=0 ⇕ 1x^2+( - 1)x+( - 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

NegNeg

- (- a)=a

x=1±sqrt((- 1)^2-4(1)(- 12))/2(1)

CalcPow

Calculate power

x=1±sqrt(1-4(1)(- 12))/2(1)

IdPropMult

Identity Property of Multiplication

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 negative and positive 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.