5. QUADRATIC EQUATIONS What do we learn in this module ? What are Quadratic Equations ? Standard form of Quadratic Equations Discriminants and their.

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What do we learn in this module ? What are Quadratic Equations ? Standard form of Quadratic Equations Discriminants and their roots Why Quadratic Equations ?

Introduction to Quadratic Equations First degree equations have variables raised to the power of 1 (one degree), as shown in the graph, and have “only one root”

Examples of first degree equations Ex. Perimeter of a square If x is the side of a square, and if the perimeter is 16 units, Perimeter = 4. x 16 = 4. x x = 4 units

Area of a square Area = x 2 Area = 4 * 4 = 16 sq.units

Example of solving quadratic equations :

Definition of a Standard Quadratic Equation

Standard form of a Quadratic Equation

Derivation of the standard equation This is called Sridhara’s method

Examples of solving using standard equation

Reducing to Quadratic form x 4 − 16x 2 − 225 = 0 x 2 = t t 2 – 16t – 225 = 0

Discriminant of a Quadratic Equation is called a discriminant >0, there are 2 unequal real solutions. =0, there is a repeated real solution. <0, there is no real solution.

The sum of the squares of 2 consecutive positive even numbers is 580. Find the numbers Identify the unknown: Let one number be x, therefore 2 nd number is x + 2 Form the equation Solve! Are both answers acceptable? (rej) :The numbers are 16 and 18Ans Statement Problems

The length and breadth of a rectangle are (3x + 1) and (2x – 1) cm respectively. If the area of the rectangle is 144 cm 2, find x. Identify the unknown! Form the equation! Solve! Are both answers acceptable? (rej)

The sum of the squares of 2 consecutive positive even numbers is 580. Find the numbers. Identify the unknown: Let one number be x, therefore 2 nd number is x + 2 Form the equation Solve! Are both answers acceptable? (rej) :The numbers are 16 and 18Ans

The perimeter of a rectangle is 44 cm. The area of the rectangle is 117 cm 2. Find the length of the shorter side of the rectangle. Let one side be x, therefore other side is (44 − 2x) ÷ 2 = 22 – x Are both answers acceptable? (rej) : The shorter side is 9 cmAns x x

A rectangular swimming pool measures 25 m by 6 m. It is surrounded by a path of uniform width. If the area of the path is 102 m 2, find the width of the path. Let the width be x. Therefore, length of path = 25 + 2x, breadth of path = 6 + 2x 25 m 6 m 25 + 2x 6 + 2x Area of pool = 25 x 6 = 150 m 2 Ans: The width of the path is 1.5 m

A duck dives under water and its path is described by the quadratic function y = 2x 2 -4x, where y represents the position of the duck in metres and x represents the time in seconds. a. How long was the duck underwater? The duck is no longer underwater when the depth is 0. We can plug in y= 0 and solve for x. So x = 0 or 4 The duck was underwater for 4 seconds

A duck dives under water and its path is described by the quadratic function y = 2x 2 -4x, where y represents the position of the duck in metres from the water and x represents the time in seconds. b. When was the duck at a depth of 5m? We can plug in y= -5 and solve for x. We cannot solve this because there’s a negative number under the square root. We conclude that the duck is never 5m below the water.

A duck dives under water and its path is described by the quadratic function y = 2x 2 -4x, where y represents the position of the duck in metres from the water and x represents the time in seconds. b. When was the duck at a depth of 5m? We can check this by finding the minimum value of y. We conclude that the duck is never 5m below the water.

A duck dives under water and its path is described by the quadratic function y = 2x 2 -4x, where y represents the position of the duck in metres and x represents the time in seconds. c. How long was the duck at least 0.5m below the water’s surface? We can plug in y= -0.5 and solve for x. The duck was 0.5m below at t = 0.14s and at t = 1.87s This will give us the times when the duck is at 0.5 m below. Therefore it was below 0.5m for 1.73s

Example f(x) = x 2 - 4 Solutions are -2 and 2. Solving using Graphical method

f(x) = 2x - x 2 Solutions are 0 and 2.

One method of graphing uses a table with arbitrary x-values. Graph y = x 2 - 4x Roots 0 and 4, Vertex (2, -4), Axis of Symmetry x = 2 xy 00 1-3 2-4 3-3 40

Why Quadratic Equations ?? http://www.youtube.com/watch?v=BjbyqgUEbAE Balls, Arrows, Missiles and Stones If you throw a ball (or shoot an arrow, fire a missile or throw a stone) it will go up into the air, slowing down as it goes, then come down again... and a Quadratic Equation tells you where it will be!Quadratic Equation

Quadratic Equations are useful in many other areas: Quadratic equations are also needed when studying lenses and curved mirrors. And many questions involving time, distance and speed need quadratic equations. I am pretty sure that economists need to use quadratic equations, too!