Exercise 1 Page 586

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. - 3x^2-11x+4=0 ⇕ - 3x^2+( - 11)x+ 4=0 We see that a= - 3, b= - 11, and c= 4. Let's substitute these values into the Quadratic Formula.

x=- b±sqrt(b^2-4ac)/2a

SubstituteValues

Substitute values

x=- ( - 11)±sqrt(( - 11)^2-4( - 3)( 4))/2( - 3)

▼

Solve for x and Simplify

NegNeg

- (- a)=a

x=11±sqrt((- 11)^2-4(- 3)(4))/2(- 3)

CalcPow

Calculate power

x=11±sqrt(121-4(- 3)(4))/2(- 3)

MultNegNegOnePar

- a(- b)=a* b

x=11±sqrt(121+12(4))/2(- 3)

Multiply

Multiply

x=11±sqrt(121+48)/- 6

AddTerms

Add terms

x=11±sqrt(169)/- 6

CalcRoot

Calculate root

x=11 ± 13/- 6

The solutions for this equation are x= 11 ± 13- 6. Let's separate them into the positive and negative cases, then round them to the nearest hundredth.

x=11 ± 13/- 6
x_1=11+13/- 6	x_2=11-13/- 6
x_1=24/- 6	x_2=- 2/- 6
x_1=- 24/6	x_2=1/3
x_1=- 4	x_2 ≈ 0.33

Using the Quadratic Formula, we found that the solutions of the given equation are x_1=- 4 and x_2 ≈ 0.33.