Exercise 1 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 We first need to identify the values of a, b, and c. x^2-2x-15=0 ⇕ 1x^2+( - 2)x+( - 15)=0 We see that a= 1, b= - 2, and c= - 15. Let's substitute these values into the Quadratic Formula.

x=- b±sqrt(b^2-4ac)/2a

SubstituteValues

Substitute values

x=- ( -2)±sqrt(( - 2)^2-4( 1)( - 15))/2( 1)

▼

Solve for x and Simplify

NegNeg

- (- a)=a

x=2±sqrt((- 2)^2-4(1)(- 15))/2(1)

CalcPow

Calculate power

x=2±sqrt(4-4(1)(- 15))/2(1)

MultByOne

a * 1=a

x=2±sqrt(4-4(- 15))/2

MultNegNegOnePar

- a(- b)=a* b

x=2±sqrt(4+60)/2

AddTerms

Add terms

x=2±sqrt(64)/2

CalcRoot

Calculate root

x=2± 8/2

FactorOut

Factor out 2

x=2(1± 4)/2

CancelCommonFac

Cancel out common factors

x= 1 ± 4

The solutions for this equation are x= 4± 22. Let's separate them into the positive and negative cases.

x=1 ± 4
x_1=1 - 4	x_2=1 + 4
x_1=- 3	x_2=5

Using the Quadratic Formula, we found that the solutions of the given equation are x_1= - 3 and x_2= 5.