for reference only

1. Quadratic equation by four different methods
E.g.1
The sum of two numbers is 27 and their product is 50. Find the numbers.
Let one number be x. Then the other number is 50/x.
x + 50/x = 27
X x => x2
+ 50 = 27x
- 27x => x2
- 27x + 50 = 0
(x -25)(x -2) = 0
(x -25) = 0 or (x -2) = 0
x = 25 or x = 2.
E.g.2
The length of a rectangle is 5 cm more than its width and the area is 50cm2
.
Find the length, width and the perimeter.
Let the width be x. Then the length = x + 5.
x(x + 5) = 50
x2
+ 5x = 50
-50 => x2
+ 5x - 50 = 0
(x + 10)(x -5) =0
(x + 10) = 0 or (x -5) =0
x = -10 or x = 5 - x = -10 is impossible to be a width
Width = 5cm; so, the length = 10cm.
Perimeter = 30cm.
E.g.3
The three sides of a right-angled triangle are x, x+1 and 5. Find x and the area,
if the longest side is 5.
The hypotenuse = 5
x2
+ (x+1)2
= 52
(Pythagoras' Theorem)
x2
+ x2
+ 2x + 1 = 25
-25 => x2
+ x2
+ 2x - 24 = 0
2x2
+ 2x - 24 = 0
x2
+ x - 12 = 0
(x - 3)(x + 4) = 0
(x + 4) = 0 or (x - 3) = 0
x = -4 or x = 3
x = 3;
Area = 1/2 x 3 x 4 = 6cm2

2. E.g.4
The product of two numbers is 24 and the mean is 5. Find the numbers.
Let one number = x; then the other = 24/x
(x + 24/x)/2 = 5
X 2 => x + 24/x = 10
X x => x2
+ 24 = 10x
- 10x => x2
+ -10x + 24 = 0
(x - 6)(x -4) = 0
(x - 6) = 0 or (x -4) = 0
x = 6 or x =4
The numbers are 6 or 4.
E.g.5
The sum of numbers is 9. The squares of the numbers is 41. Find the numbers.
These are quadratic simultaneous equations.
let the numbers be x and y.
x + y = 9
x2
+ y2
= 41
From the first equation, y = (9-x)
Now substitute this in the second equation.
x2
+ (9-x)2
= 41
x2
+ 81 - 18x + x2
= 41
2x2
- 16x + 81 = 41
2x2
- 16x + 40 = 0
x2
- 8x + 20 = 0
(x - 5)(x -4) =0
(x - 5) = 0 or (x -4) =0
x = 5 or x = 4
Substitute in the first equation, y = 5 or 4
The numbers are 5 and 4.
E.g.6
A ball is thrown upwards from a rooftop, 80m above the ground. It will reach a
maximum vertical height and then fall back to the ground. The height of the
ball from the ground at time t is h, which is given by,
h = -16t2
+ 64t + 80.

3. 1. What is the height reached by the ball after 1 second?
2. What is the maximum height reached by the ball?
3. How long will it take before hitting the ground?
Follow the graph along with the calculation for a better understanding:
1) h = -16t2
+ 64t + 80
h = -16* 1*1 + 64*1 + 80 = 128m
2) Rearrange by the completing the square, we get:
h = -16[t2
- 4t - 5]
h = -16[(t - 2)2
- 9]
h = -16(t - 2)2
+ 144
When the height is maximum, t = 2; therefore, maximum height = 144m.
3) When the ball hits the ground, h = 0;
-16t2
+ 64t + 80 = 0
Divide the equation by -16
t2
- 4t - 5 = 0

4. (t - 5)(t + 1) = 0
t = 5 or t = -1
The time cannot be negative; so, the time = 5 seconds.
E.g.7
Two resistors, when connected in series, have a total resistance of 25 Ohms. If
they are connected in parallel, the value goes down to 6 Ohms. Find the
values.
When they are in series, if one resistor is x, then the other is 25-x
When they are in parallel, 1/6 = 1/x + 1/(25-x)
1/6 = 25/[x(25-x)] = 25/[25x - x2
]
25x - x2
= 150
If ax2
+ bx + c = 0, then
x = [-b ±√(b2
- 4ac) ]/ 2a
x2
- 25x + 150 = 0
a = 1; b = -25; c = 150
x = -(-25) ±√((-25)2
- 4(1)(150)) / 2(1)
x = 25 ±√(625 - 600) / 2
x = 25 ±√(25) / 2
x = (25 ± 5 )/ 2
x = 15 or x = 10
So, the resistors are 15 Ohms or 10 Ohms.
E.g.8
A farmer wants to make a rectangular pen for his sheep. He has 60m fencing
material to cover three sides with the other side being a brick wall. How should
he use the fencing material to maximize the space for his sheep? How should
he choose length and width of the pen to achieve his objective?

5. He just has to cover three sides; let the width be x.
Then the length = (60-2x)
Area of the pen = x(60-2x)
= 60x -2x2
Now let's sketch a graph for the quadratic equation. Which is as follows:

6. As you can see, the curve peaks at x = 15; when the width = 15m, the area is
maximum. A very useful way to use quadratic equations in real life, indeed!
E.g.9
The following picture shows the shape of a certain grass patch. If the area of
the patch is 80m2
, find k.
The total area = 5k + k(2k+1)
= 5k + 2k2
+ k
= 2k2
+ 6k
Since the area is 80m2
2k2
+ 6k = 80
2k2
+ 6k - 80 = 0
(2k - 10)(k + 8) = 0
k = 5 or k = -8
Since the length cannot be negative, k = 5.
E.g.10
The following picture shows the shape of a rectangle from which a smaller
rectangular part is removed. If the remaining area of the larger rectangle is
35cm2
, find k.

7. The remaining area = k(2k+6) - 3k
= 2k2
+ 6k -3k
= 2k2
+ 3k
Since the area is 35cm2
2k2
+ 3k = 35
2k2
+ 3k - 35 = 0
(2k -7)(k + 5) = 0
k = 3.5 or k = -5
Since the length cannot be negative, k = 3.5.
E.g.11
The shortest side of a right-angled triangle is 6cm shorter than its hypotenuse.
The difference in length of other two sides is 3cm. If the shortest side is n-3,
show that 2n2
= 12n. Hence, find n.

8. If the length of the shortest side is n-3, the length of the hypotenuse and the
other side are n+3 and n respectively.
So, using Pythagoras Theorem,
(n-3)2
+ n2
= (n+3)2
n2
- 6n + 9 + n2
= n2
+ 6n + 9
2n2
= 12n
2n2
- 12n = 0
2n2
= 12n
n2
- 6n = 0
n = 0 or n = 6.
Since the length cannot be zero, n = 6.