University:
Missouri State University
Course:
MTH 103 | Intermediate Algebra
Academic year:

2020
Views:

333

Pages:

2
Author:

Sydnee Rodriguez

Algebra 2 Trig Name: KEY Date: Section 5.6-Solving Quadratic Equations by the Quadratic Formula The Quadratic Formula: x = (- b plus/minus sqrt(b ^ 2 - 4ac))/(2a) Before you can use the quadratic formula to solve an equation, the equation must be in standard form: a * x ^ 2 + bx + c = 0 How to solve a quadratic equation using the Quadratic Formula: Examples: 1) 3x ^ 2 + 8x = 35 3x ^ 2 + 8x - 35 = 0 x = (- 8 plus/minus sqrt(8 ^ 2 - 4(3)(- 35)))/(2(3)) x= -8 pm boxed 6A + 420 6 x = (- 8 plus/minus sqrt(484))/(- 46mu) x = (- 7 plus/minus 22)/163 x = (- 4 plus/minus 11)/3 -> x = 7/3, - 5 2) 2x ^ 2 + x = 5 2x ^ 2 + x - 5 = 0 x = (- 1 plus/minus sqrt(1 ^ 2 - 4(2)(- 5)))/(2/2) x = (- 1 plus/minus sqrt(1 + 40))/4 x_{0} = (- 1 plus/minus sqrt(41))/4 2 Real Solutions (2 x-intercepts) 2 Real Solutions (2 x-intercepts) 3) x ^ 2 - x = 5x - 9 x ^ 2 - 6x * 9 = 0 x = (6 plus/minus sqrt((- 6) ^ 2 - 4(1)(9)))/21 x = (6 plus/minus sqrt(36 - 36))/2 x = (6 plus/minus sqrt(0))/2 x = 6/2 x = 3 1 Real Solution (1 x-intercept, the vertex) 4) - x ^ 2 + 2x = 2 - x ^ 2 + 2x - 2 = 0 x = (- 2 plus/minus sqrt(2 ^ 2 - 4(- 1)(- 2)))/(2(- 1)) x = (- 2 plus/minus sqrt(4 - 8))/- 2 x= - 2 plus/minus sqrt(- 4) 1^ -2 + x = - 1/(x + 1/7) * 7 x=1 pm-j |x=1 pm j 2 imaginary Solutions (no xintercepts) You can determine the number of solutions (1 or 2) and the type of solutions (real or imaginary) that a quadratic equation has without using all of the quadratic formula. x = (- b plus/minus sqrt(b ^ 2 - 4ac))/(2a) The Discriminant: b ^ 2 - 4ac If b ^ 2 - 4ac is positive 2 Real Solutions . If b ^ 2 - 4ac is zero 1 Real Solution If b ^ 2 - 4ac is negative 2 Imaginary Solutions **Remember, the quadratic equation must be in standard form! a * x ^ 2 + bx + c = 0 Example: Find the discriminant of the equation and give the number and type of solutions. Then find the solutions! 1) x ^ 2 - 6x + 8 = 0 Discriminant: (- 6) ^ 2 - 4(1)(8) 36-32 4 2 Real Solutions Solutions: x= ( 6 plus/minus sqrt((- 6) ^ 2 - 4(1)(8)) 2(1) = 6 pm sqrt 30-32 2 x= 6 pm sqrt 4 2 = root(6 plus/minus y', 3) * x_{1} = 3 plus/minus 1 = boxed 4,2 2) 9x ^ 2 + 6x + 3 = 2 9x ^ 2 + 6x + 1 = 0 6 ^ 2 - 4(9)(1) 0 1 Real Solution 3) 5x ^ 2 + 5x - 2 = 2x - 3 5x ^ 2 + 3x + 1 = 0 Discriminant: 3 ^ 2 - 4(5)(1) 9-20 2 imaginary Solutions Solution: x = (- 6 plus/minus sqrt(6 ^ 2 - 4(9 * 1)))/(2(9)) = (- 6 plus/minus sqrt(36 - 36))/18 x = (- 6 plus/minus sqrt(0))/18 = - 6/18 = boxed -1 3 Solutions: x = (- 3 plus/minus sqrt(3 ^ 2 - 4(5)(1)))/(2(5)) = (- 3 plus/minus sqrt(9 - 20))/10 x = (- 3 plus/minus sqrt(- 11))/10 = boxed -3 pm i sqrt 11 10