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Solving Quadratic Functions Lesson 5-3 Objective • Today, you will . . . • solve quadratic functions by using a variety of methods. TEKS:b2A,d1A,d3A,d3C,d3D Some Notes on Quadratic Functions 1. The graphs of quadratic functions are parabolas. 2. The solution(s) that you will be looking for are the x-intercepts of the parabola. 3. The x-intercepts are also called the “roots,” “solutions,” or the “zeros” 4. Quadratic functions can have one, two, or no real solutions. 5. Quadratic functions that have no real solutions have complex (imaginary) solutions. Different Graphs of Quadratics Two real solutions X=-6 X=1 x-intercepts, roots, zeros Different Graphs of Quadratics One real solutions X=3 x-intercept, root, zero Different Graphs of Quadratics No real solutions No Solution No x-intercepts (roots or zeros) Today you’ll find… • The solutions to quadratic equations by factoring • For example: GIVEN: y = Ax2 + Bx + C FIND: The solutions, roots, zeros, or x-intercepts Solve by factoring: x2 + 9x + 20 = 0 1 x 20 x2+ 9x + 20 = 0 2 x 20 4x5 (x + 5)(x + 4) = 0 x+5=0 x+4=0 x=-5 x=-4 So, its two roots, solutions, zeros are -5 & -4 Solve by factoring: x2 = -7x + 18 x2 + 7x - 18 = 0 1 x 18 2x9 Hint: The sign of “B” goes with 3x6 the largest factor! (x - 2)(x + 9) = 0 x-2=0 x+9=0 x=2 x=-9 So, its two roots, solutions, zeros are 2 & -9 Solve by factoring: x2 + 100 = 29x 1 x 100 x2 - 29x + 100 = 0 2 x 50 4 x 25 5 x 20 (x - 4)(x - 25) = 0 10 x 10 x-4=0 x - 25 = 0 x=4 x = 25 So, its two roots, solutions, zeros are 4 & 25 Solve by factoring: x2 - 9 = 0 1x9 x2 -9=0 3x3 (x + 3)(x - 3) = 0 x+3=0 x-3=0 x=-3 x=3 So, its two roots, solutions, zeros are -3 & 3 Solve by factoring: x2 + x = 6 1x6 x 2 +x-6=0 2x3 Hint: The sign of “B” goes with the largest factor! (x - 2)(x + 3) = 0 x-2=0 x+3=0 x=2 x=-3 So, its two roots, solutions, zeros are 2 & -3 Solve by factoring: x2 - 6x = - 8 1x8 x=2 2x4 x=4 Solve by factoring: x2 - 8 = - 7x x=1 1x8 2x4 x=-8 Solve by factoring: 2x2 + 13x + 15 = 0 2x2 + 13x + 15 = 0 Multiply AxC = 30 Determine the factors of 30 that 1 x 30 give you 13 2 x 15 3 x 10 (x + 3) (x + 10) Write the factors (x + 3) (x + 10 ) Divide the #’s by A 2 2 (2x + 3) (x + 5) If not divisible, send it in front of “x”, if divisible then simplify. 2x + 3 = 0 x+5=0 Now solve both factors! 2x = -3 x = -5 Your solutions are x = -3/2 x = -3/2 and x = -5 Solve by factoring: 3x2 + 16x + 21 = 0 3x2 + 16x + 21 = 0 Multiply AxC = 63 Determine the factors of 63 that 1 x 63 give you 16 3 x 21 7x9 (x + 7) (x + 9) Write the factors (x + 7) (x + 9 ) Divide the #’s by A 3 3 (3x + 7) (x + 3) If not divisible, send it in front of “x”, if divisible then simplify. 3x + 7 = 0 x+3=0 Now solve both factors! 3x = -7 x = -3 Your solutions are x = -7/3 x = -7/3 and x = -3 Solve by factoring: 2x2 – 5x = 7 2x2 – 5x – 7 = 0 Multiply AxC = -14 Determine the factors of -14 that 1 x 14 give you -5 2x7 (x + 2) (x – 7) Write the factors Remember, sign of “B” goes to the largest factor, in this case, the negative goes to the 7. (x + 2) (x – 7) Divide the #’s by A 2 2 (x + 1) (2x – 7) If not divisible, send it in front of “x”, if divisible then simplify. x+1=0 2x – 7 = 0 Now solve both factors! x = -1 2x = 7 Your solutions are x = 7/2 x = -1 and x = 7/2 Your turn! Solve by factoring: 5x2 + 4x = 12 5x2 + 4x – 12 = 0 AxC= -60 1 x 60 4 x 15 2 x 30 5 x 12 3 x 20 6 x 10 (x – 6) (x + 10) Write the factors (watch your signs!) (x – 6) (x + 10) Divide the #’s by A 5 5 (5x – 6)(x + 2) If not divisible, send it in front of “x”, if divisible then simplify. 5x – 6 = 0 x+2=0 Now solve both factors! 5x = 6 x = -2 Your solutions are x = 6/5 x = 6/5 and x = -2 You try these by factoring . . . 1. 2x2 – 3x = 0 x = 0 and 3/2 2. 4x2 + 5 = -9x x = -1 and -5/4 3. 6x2 + 55x = -9 x = -9 and -1/6

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