Problem Solving With Quadratic Equations x 2 8

Problem Solving With Quadratic Equations

x 2 + 8 x + 16 = 0 Graphically Graph related function y = x 2 + 8 x + 16 Algebraically x 2 + 8 x + 16 = 0 x 2 + 4 x+ 16 = 0 (x + 4)(x+ 4) = 0 x + 4 = 0 or x+ 4 = 0 x = -4 or x = -4 Two equal real number roots.

Solving with Complex Bases (a + 2)2 + 3(a + 2) + 2 = 0 Let B = (a + 2) Replace (a + 2) with B. [(a + 2) + 2] [(a + 2) + 1] = 0 (a + 4)(a + 3) = 0 (a + 4) = 0 or (a + 3) = 0 a = -4 or a = -3 B 2 + 3 B + 2 = 0 (B + 2)(B + 1) = 0

Writing a Quadratic Equation With Given Roots Write a possible quadratic equation, given the following roots: x = -6 or x = 3 The factors are (x + 6) and (x - 3). (x + 6)(x - 3) = 0 x 2 + 3 x - 18 = 0 Is this the only equation?

Solve a Problem Involving Quadratic Equations height of jet of water nozzle horizontal distance light Suppose the water fountain jet creates a parabolic stream of water modelled by the quadratic function where h is the height of the jet of water and x is the horizontal distance of the jet of water from the water nozzle, both in metres. How far from the water nozzle should the underwater light be placed for the jet of water to land on it? Solve the equation: or The distance between the x-intercepts of the graph is 2. 5 metres. The underwater light must be placed 2. 5 metres from the nozzle.

The product of two consecutive even integers is 224. Determine the value of the integers. Let x = first even integer Let x + 2 = second even integer x(x + 2) = 224 x 2 + 2 x - 224 = 0 x = - 16 or x = 14 Note there are two possible solutions for x. Therefore, there are two sets of answers. When x = -16, the integers are -16 and -14. When x = 14, the integers are 14 and 16.

1 Find the lengths of the two unknown sides of a triangle if the hypotenuse is 15 cm long and the sum of other two legs is 21 cm. Let x = one side of the triangle 21 - x will be the other side. x 2 + (21 - x)2 = (15)2 x 2 + 441 - 42 x + = 225 2 x 2 - 42 x + 441 = 225 2 x 2 - 42 x + 216 = 0 2(x 2 - 21 x + 108) = 0 2(x - 9)( x - 12) = 0 x = 9 or x = 12 Therefore, the lengths of the sides of the triangle are 9 cm x 12 cm.

2. 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? x Let x = the width of the strip. x 60 - 2 x 80 - 2 x x 80 Total area = 80 x 60 = 4800 m 2 Area of the factory: = 2400 m 2 x 60 Area of the factory: 2400 = (80 - 2 x)(60 - 2 x) 2400 = 4800 - 280 x + 4 x 2 0 = 4 x 2 - 280 x + 2400 0 = 4(x 2 - 70 x + 600) 0 = 4(x - 60)(x - 10) x = 60 or x = 10 extraneous Therefore, the strip is 10 m wide. The factory is 60 m x 40 m.

3 A picture that measures 30 cm by 20 cm is to be surrounded by a frame of uniform width and equal in area to the picture. How wide is the frame, and what are the dimensions of the entire framed picture? x Let x = the width of the frame x 30 20 x x 30 + 2 x Total area = 30 x 2 = 1200 cm 2 20 + 2 x 1200 = (30 + 2 x)(20 + 2 x) 1200 = 600 + 100 x + 4 x 2 0 = 4 x 2 + 100 x - 600 0 = 4(x 2 + 25 x - 150) 0 = 4(x - 5)(x + 30) x = 5 or x = -30 extraneous Therefore, the frame is 5 cm wide. The framed picture is 30 cm x 40 cm.

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