Exercise 30 Page 967

We want to find all the complex solutions of the given quadratic equation. To do it, we will use the Quadratic Formula. 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^2-4x=- 5 ⇔ x^2-4x+5=0Now, we can identify the values of a, b, and c. x^2-4x+5=0 ⇕ 1x^2+( - 4)x+ 5=0 We see that a= 1, b= - 4, and c= 5. Let's substitute these values into the Quadratic Formula.

x=- b±sqrt(b^2-4ac)/2a

SubstituteValues

Substitute values

x=- ( -4)±sqrt(( - 4)^2-4( 1)( 5))/2( 1)

NegNeg

- (- a)=a

x=4±sqrt((- 4)^2-4(1)(5))/2(1)

CalcPow

Calculate power

x=4±sqrt(16-4(1)(5))/2(1)

Multiply

Multiply

x=4±sqrt(16-20)/2

SubTerm

Subtract term

x=4±sqrt(- 4)/2

Since we have a negative radicand, we can recall the definition of the imaginary unit i. sqrt(- 1) = i We will use this definition to continue simplifying.

x=4±sqrt(- 4)/2

SqrtNegToSqrtI

sqrt(- a)= sqrt(a)* i

x=4± sqrt(4)i/2

CalcRoot

Calculate root

x=4± 2i/2

FactorOut

Factor out 2

x=2(2± i)/2

CancelCommonFac

Cancel out common factors

x=2(2± i)/2

SimpQuot

Simplify quotient

x=2± i

Using the Quadratic Formula, we found that the complex solutions are x_1=2+i and x_2=2-i. This corresponds to option C.