The Discriminant 4.7A Chapter 4 Quadratic and Polynomial Equations MATHPOWERTM 11, WESTERN EDITION 4.7.1

will give the roots of the quadratic equation. The Nature of the Roots The quadratic formula will give the roots of the quadratic equation. From the quadratic formula, the radicand, b2 - 4ac, will determine the Nature of the Roots. By the nature of the roots, we mean: whether the equation has real roots or not if there are real roots, whether they are different or equal The expression b2 - 4ac is called the discriminant of the equation ax2 + bx + c = 0 because it discriminates among the three cases that can occur. 4.7.2

x2 + x - 6 = 0 The Nature of the Roots [cont’d] The value of the discriminant is greater than zero. x = -3 or 2 There are two distinct real roots. 4.7.3

The Nature of the Roots [cont’d] x2 + 2x + 1 = 0 The value of the discriminant is 0. There are two equal real roots. x = -1 or - 1 4.7.4

The Nature of the Roots [cont’d] 2x2 - 3x + 8 = 0 The value of the discriminant is less than zero. Imaginary roots 4.7.5

The Nature of the Roots [cont’d] We can make the following conclusions: If b2 - 4ac > 0, then there are two different real roots. If b2 - 4ac = 0, then there are two equal real roots. If b2 - 4ac < 0, then there are no real roots. Determine the nature of the roots: x2 - 2x + 3 = 0 4x2 - 12x + 9 = 0 b2 - 4ac = (-2)2 - 4(1)(3) = -8 b2 - 4ac = (-12)2 - 4(4)(9) = 0 There are two equal real roots. The equation has no real roots. 4.7.6

Determine the value of k for which the equation x2 + kx + 4 = 0 has a) equal roots b) two distinct real roots c) no real roots a) For equal roots, b2 - 4ac = 0. Therefore, k2 - 4(1)(4) = 0 k2 - 16 = 0 k2 = 16 k = + 4 The equation has equal roots when k = 4 and k = -4. b) For two different real roots, b2 - 4ac > 0. k2 - 16 > 0 k2 > 16 Therefore, k > 4 or k< -4. This may be written as | k | > 4. Which numbers when squared produce a result that is greater than 16? Which numbers when squared produce a result that is less than 16? c) For no real roots, b2 - 4ac < 0. k2 < 16 Therefore, -4 < k < 4. This may be written as | k | < 4. 4.7.7

Assignment Suggested Questions: Page 189 1, 2, 3, 18, 22, 28-30, 33, 34, 38 Read page 154, example 5. Page 156 44, 51 4.7.8

Quadratic Equations - Word Problems Chapter 4 Quadratic and Polynomial Equations 4.7B Quadratic Equations - Word Problems MATHPOWERTM 11, WESTERN EDITION 4.7B.9

1. The product of two consecutive even integers is 224. Find the integers. Let x = 1st even integer Let y = 2nd even integer y = x + 2 xy = 224 x(x + 2) = 224 x2 + 2x = 224 x2 + 2x - 224 = 0 (x + 16)(x - 14) = 0 x = - 16 or x = 14 When x = -16, the integers are -16 and -14. When x = 14, the integers are 14 and 16. Note there are two possible solutions for x. Therefore, there are two sets of answers. 4.7B.10

2. The side of one square is 3 cm longer than the side of another. If their combined area is 117 cm2, find the dimensions of each square. y = x + 3 Let x = side of the smaller square. Let y = side of the larger square. x2 + y2 = 117 Therefore, the smaller square is 6 cm x 6 cm. x2 + (x + 3)2 = 117 x2 + x2 + 6x + 9 =117 2x2 + 6x + 9 = 117 2x2 + 6x - 108 = 0 2(x2 + 3x - 54) = 0 2(x + 9)( x - 6) = 0 x = -9 or x = 6 The larger square is 9 cm x 9 cm. Not possible 4.7B.11

3. A factory is to be built on a lot that measures 80 m by 60 m. A lawn of uniform width and equal in area to the factory, must surround the factory. How wide is the strip of lawn, and what are the dimensions of the factory? Let x = the width of the strip. x Area of the factory: x x 60 - 2x 60 2400 = (80 - 2x)(60 - 2x) 2400 = 4800 - 280x + 4x2 0 = 4x2 - 280x + 2400 0 = 4(x2 - 70x + 600) 0 = 4(x - 60)(x - 10) x = 60 or x = 10 80 - 2x x 80 Total area = 80 x 60 = 4800 m2 Area of the factory: Therefore, the strip is 10 m wide, and the factory is 60 m x 40 m. = 2400 m2 4.7B.12

+ = 5 4. Kelly drives 230 km in 5 h from her home in Whitehorse. For the last 150 km of the trip, she increases her average speed by 10 km/h. What is the average speed for each distance traveled? d s t d = st Slower 80 x Let x = slower speed x + 10 150 Faster 80x + 800 + 150x = 5x2 + 50x 5x2 - 180x - 800 = 0 5(x2 - 36x - 160) = 0 5(x - 40)(x + 4) = 0 x = 40 or x = -4 Slower time + Faster time = Total time x(x + 10) + = 5 80(x + 10) + 150x = 5(x)(x + 10) Therefore, the slower speed was 40 km/h and the faster was 50 km/h. 4.7B.13