Exercise 56 Page 627

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^2-20=8x

SubEqn

LHS-8x=RHS-8x

x^2-8x-20=0

Now, we can identify the values of a, b, and c. x^2-8x-20=0 ⇕ 1x^2+( - 8)x+( - 20)=0 We see that a= 1, b= - 8, and c= - 20. Let's substitute these values into the Quadratic Formula.

x=- b±sqrt(b^2-4ac)/2a

SubstituteValues

Substitute values

x=- ( -8)±sqrt(( - 8)^2-4( 1)( - 20))/2( 1)

▼

Solve for x and Simplify

NegNeg

- (- a)=a

x=8±sqrt((- 8)^2-4(1)(- 20))/2(1)

CalcPow

Calculate power

x=8±sqrt(64-4(1)(- 20))/2(1)

IdPropMult

Identity Property of Multiplication

x=8±sqrt(64-4(- 20))/2

MultNegNegOnePar

- a(- b)=a* b

x=8±sqrt(64+80)/2

AddTerms

Add terms

x=8±sqrt(144)/2

CalcRoot

Calculate root

x=8± 12/2

The solutions for this equation are x= 8± 122. Let's separate them into the positive and negative cases.

x=8± 12/2
x_1=8+12/2	x_2=8-12/2
x_1=20/2	x_2=- 4/2
x_1=10	x_2=- 2

Using the Quadratic Formula, we found that the solutions of the given equation are x_1=10 and x_2=- 2. Therefore, the answer is J.