5. QUADRATIC EQUATIONS
What do we learn in this module ?
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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 = x2
Area = 4 * 4 = 16 sq.units
Examples of quadratic equations
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
x4 − 16x2 − 225 = 0
x2 = t
t2 – 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.
Statement Problems
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 2nd number is x + 2
x2  ( x  2)2  580
x 2  x 2  4 x  4  580
Form the equation
2 x 2  4 x  576  0
x 2  2 x  288  0
( x  16)( x  18)  0
x  16 or  18 (rej)
Are both answers acceptable?
Solve!
Ans :The numbers are 16 and 18
The length and breadth of a rectangle are
(3x + 1) and (2x – 1) cm respectively. If the area
of the rectangle is 144 cm2, find x.
(3x  1)(2 x  1)  144
6 x  3x  2 x  1  144
2
Identify the unknown!
Form the equation!
6 x 2  x  145  0
(6 x  29)( x  5)  0
Solve!
29
x  5 or  (rej)
6
Are both answers acceptable?
Ans : x  5
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 2nd number is x + 2
x2  ( x  2)2  580
x 2  x 2  4 x  4  580
Form the equation
2 x 2  4 x  576  0
x 2  2 x  288  0
( x  16)( x  18)  0
x  16 or  18 (rej)
Are both answers acceptable?
Solve!
Ans :The numbers are 16 and 18
The perimeter of a rectangle is 44 cm. The area
of the rectangle is 117 cm2. Find the length of
the shorter side of the rectangle.
Let one side be x, therefore other side is (44 − 2x) ÷ 2
= 22 – x
x(22  x)  117
22 x  x 2  117
x
x 2  22 x  117  0
( x  9)( x  13)  0
x  9 or 13 (rej)
Are both answers acceptable?
Ans : The shorter side is 9 cm
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 m2, find the width of the path.
Let the width be x. Therefore, length of path = 25 + 2x,
breadth of path = 6 + 2x
Area of pool = 25 x 6 = 150 m2
(25  2 x)(6  2 x)  252
150  50 x  12 x  4 x 2  252
4 x 2 62 x  102  0
25 + 2x
6 + 2x
25 m
6m
2 x 2  31x  51  0
Ans: The width of the path is 1.5 m
( x  17)(2 x  3)  0
x  1.5 or -17(rej)
A duck dives under water and its path is described by
the quadratic function y = 2x2 -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.
0  2x2  4x
0  2 x( x  4)
0  2x
0  x4
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 = 2x2 -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.
 5  2x2  4x
0  2x2  4x  5
 b  b 2  4a c
x
2a
 (4)  (4) 2  4(2)(5)
x
2( 2)
4  1 6 4 0
4
4 24
x
4
x
A duck dives under water and its path is described by
the quadratic function y = 2x2 -4x, where y represents
the position of the duck in metres from the water and
x represents the time in seconds.
We conclude that
the duck is never
5m below the
water.
b. When was the duck at a depth of 5m?
We can check this by finding the
minimum value of y.
b
2a
 ( 4)
x 
2( 2)
4
x 
4
x 1
x 
y  2(1) 2  4(1)
y  2
A duck dives under water and its path is described by
the quadratic function y = 2x2 -4x, where y represents
the position of the duck in metres and x represents
the time in seconds.
 0 .5  2 x 2  4 x
0  2 x 2  4 x  0 .5
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.
This will give us the
times when the duck is
at 0.5 m below.
The duck was 0.5m
below at t = 0.14s
and at t = 1.87s
Therefore it was
below 0.5m for 1.73s
 b  b 2  4ac
x
2a
 (4)  (4) 2  4(2)(0.5)
x
2(2)
4  16  4
4
4  12
x
4
4  3.46
x
4
x  0.14 or 1.87 s
x
Solving using Graphical method
2
Example f(x) = x - 4
4
2
-5
5
-2
-4
Solutions are -2 and 2.
2
f(x) = 2x - x
4
2
Solutions are 0 and 2.
5
-2
-4
One method of graphing uses a table with arbitrary
x-values.
Graph y = x2 - 4x
4
x
0
1
2
3
4
y
0
-3
-4
-3
0
2
5
-2
Roots 0 and 4 , Vertex (2, -4) ,
Axis of Symmetry x = 2
-4
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 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!