Exercise 48 Page 138

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 - 2x = 5 ⇔ x^2 - 2x - 5 = 0 Now, we can identify the values of a, b, and c. x^2 - 2x - 5 = 0 ⇕ 1x^2+( - 2)x+( - 5)=0 We see that a= 1, b= - 2, and c= - 5. 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)( - 5))/2( 1)

▼

Solve for x and Simplify

NegNeg

- (- a)=a

x=2±sqrt((- 2)^2-4(1)(- 5))/2(1)

CalcPow

Calculate power

x=2±sqrt(4-4(1)(- 5))/2(1)

MultByOne

a * 1=a

x=2±sqrt(4-4(- 5))/2

MultNegNegOnePar

- a(- b)=a* b

x=2±sqrt(4+20)/2

AddTerms

Add terms

x=2±sqrt(24)/2

SplitIntoFactors

Split into factors

x=2±sqrt(4* 6)/2

SqrtProd

sqrt(a* b)=sqrt(a)*sqrt(b)

x=2± sqrt(4)* sqrt(6)/2

CalcRoot

Calculate root

x=2 ± 2 sqrt(6)/2

FactorOut

Factor out 2

x=2(1± sqrt(6))/2

CancelCommonFac

Cancel out common factors

x=1 ± sqrt(6)

The solutions for this equation are x=1 ± sqrt(6). Let's separate them into the positive and negative cases.

x=1 ± sqrt(6)
x_1=1 + sqrt(6)	x_2=1 - sqrt(6)
x_1≈ 3.4	x_2 ≈ - 1.4

Using the Quadratic Formula, we found that the solutions of the given equation are x_1≈ 3.4 and x_2 ≈ - 1.4.