Exercise 52 Page 767

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+13x = - 36

AddEqn

LHS+36=RHS+36

x^2+13x + 36 = 0

Now, we can identify the values of a, b, and c. x^2+13x+36=0 ⇕ 1x^2+ 13x+ 36=0 We see that a= 1, b= 13, and c= 36. Let's substitute these values into the Quadratic Formula.

x=- b±sqrt(b^2-4ac)/2a

SubstituteValues

Substitute values

x=- 13±sqrt(13^2-4( 1)( 36))/2( 1)

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Solve for x and Simplify

CalcPow

Calculate power

x=- 13±sqrt(169-4(1)(36))/2(1)

Multiply

Multiply

x=- 13±sqrt(169-144)/2

SubTerm

Subtract term

x=- 13±sqrt(25)/2

CalcRoot

Calculate root

x=- 13 ± 5/2

The solutions for this equation are x= - 13 ± 52. Let's separate them into the positive and negative cases.

x=- 13± 5/2
x_1=- 13 + 5/2	x_2=- 13 - 5/2
x_1=- 8/2	x_2=- 18/2
x_1=- 4	x_2=- 9

Using the Quadratic Formula, we found that the solutions of the given equation are x_1=- 4 and x_2=- 9.