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2.4 Solving Quadratic Equations

2.4 Solving Quadratic Equations. A quadratic equation is of the form a x 2 + b x + c = 0 There are several ways to solve a quadratic 1. Factoring 2. Completing the square 3. The quadratic formula Let’s practice one of each kind!.  setting each factor = to 0. Ex 1) Factoring

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2.4 Solving Quadratic Equations

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  1. 2.4 Solving Quadratic Equations

  2. A quadratic equation is of the form ax2 + bx + c = 0 There are several ways to solve a quadratic 1. Factoring 2. Completing the square 3. The quadratic formula Let’s practice one of each kind!  setting each factor = to 0

  3. Ex 1) Factoring A company makes holiday themed stuffed animals. For October, their revenue was projected to be R(x) = 21x2 + 75x + 235 where x is the number of each 100 stuffed animals produced. The cost to make the animals is C(x) = 4x2 + 7x + 1200. They sold all the animals, but there were some production problems and the company had a loss of $200 for October. Find the number of stuffed animals they sold. Note: Profit = Revenue – Cost –200 = 21x2 + 75x + 235 – (4x2 + 7x + 1200) 0 = 17x2 + 68x –735 0 = 17(x2 + 4x –45) 0 = 17(x + 9)(x – 5) x = –9, 5 (they lost money!) *(remember to subtract WHOLE equation) so… 500 stuffed animals

  4. Ex 2) Completing the Square A small rocket is fired into the air from an initial height of 50 ft above ground level. The height, s in feet, of the rocket at any time, t in seconds, is modeled by s = –16t2 + 320t + 50. Find the time the rocket is 25 ft above the ground. (round to nearest second) s = 25 = –16t2 + 320t + 50 16t2 – 320t = 25 t2 – 20t = 1.5625 + 100 + 100 so... 20 seconds t = 20.078, –0.078

  5. 3 3 Ex 3) Quadratic Formula A rectangular open-top box is to be made by cutting 3 in. squares from each corner of a sheet of metal & folding up the edges. The length of the finished box must be 4 in. longer than twice the width. If the volume of the box must be 150 in3, determine the dimensions of the flat sheet of metal needed to construct the box. V = l•w•h h = 3 w = x l =2x + 4 V = 150 = (2x + 4)(x)(3) 150 = 6x2 + 12x 0 = 6x2 + 12x – 150 3 3 4.099 3 3 3 3 w = 4.099 + 3 + 3 = 10.099 • l = (2(4.099)+ 4) + 3 + 3 • = 18.198 • 10.099 in ⨯ 18.198 in

  6. Not all answers to a quadratic are real. What if in the quadratic formula a negative number appears under the radical?! We have to consider imaginary (use i) answers. Types of Solutions for a Quadratic • b2 – 4ac > 0 2) b2 – 4ac = 0 3) b2 – 4ac < 0 (positive # under )  2 real solutions ( …nothing to + or –)  1 real solution (negative # under ) • 0 real solutions, • 2 imag comp conj Ex 4) Determine the nature of the solutions of 6x2 – 2x + 3 = 0 (–2)2 – 4(6)(3) = 4 – 72 = – 68 < 0 2 imaginary complex conjugates

  7. (change sign in middle) Since complex (imaginary) answers come as a pair, if we know one, we know the other and can write an equation with those solutions! • conjugates: a + bi & a – bi • If a quadratic has solutions a & b, its factors are (x – a)(x – b) and the quadratic is x2 – (a + b)x + (a • b) = 0 sum product Ex 5) Write the equation of a quadratic that has 1 – i as one solution. 1st solution: 1 – i 2nd solution: 1 + i x2 – 2x + 2 = 0 sum: (1 – i) + (1 + i) = 2 product: (1 – i)(1 + i) = 1 + i – i – i2 = 1 – (–1) = 2

  8. Homework #204 Pg 84 #1–11 odd, 15–27 odd, 4, 14, 31, 33, 35, 37, 38

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