Exercise 42 Page 568

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. x^2=12x+35 ⇓ x^2-12x-35=0 Now we can identify the values of a, b, and c. x^2-12x-35=0 ⇕ 1x^2+( - 12)x+( - 35)=0 We see that a= 1, b= - 12, and c= - 35. Let's substitute these values into the Quadratic Formula.

x=- b±sqrt(b^2-4ac)/2a

SubstituteValues

Substitute values

x=- ( - 12)±sqrt(( - 12)^2-4( 1)( - 35))/2( 1)

▼

Solve for x and Simplify

NegNeg

- (- a)=a

x=12±sqrt((- 12)^2-4(1)(- 35))/2(1)

CalcPow

Calculate power

x=12±sqrt(144-4(1)(- 35))/2(1)

Multiply

Multiply

x=12±sqrt(144-4(- 35))/2

MultNegNegOnePar

- a(- b)=a* b

x=12±sqrt(144+140)/2

AddTerms

Add terms

x=12±sqrt(284)/2

SplitIntoFactors

Split into factors

x=12±sqrt(4* 71)/2

SqrtProd

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

x=12± sqrt(4)* sqrt(71)/2

CalcRoot

Calculate root

x=12± 2 sqrt(71)/2

FactorOut

Factor out 2

x=2(6± sqrt(71))/2

CancelCommonFac

Cancel out common factors

x=6± sqrt(71)

Using the Quadratic Formula, we found that the solutions of the given equation are x=6± sqrt(71). Therefore, the solutions are x_1=6+sqrt(71) and x_2=6-sqrt(71).