In This presentation we
deal with the solving of
Quadratic Equations by
splitting the middle
Once upon a time in a far away forest
there was a pious ashram where a
renowned guru named Brahmagupta and
his disciples lived and learnt new
sciences each day.
The guru used to periodically ask
his disciples brilliant and logical
questions and rewarded them
suitably if they succeeded.
On one such day he asked his
I have a problem and want you to help me out!
I have 21 goats and 12 cows and want to build an animal
shelter to keep them in.
Each goat occupies an area of 1 sq.unit and a cow
occupies double that area.
I have limited fencing of 26 units.
SO now I want you to find out what should be the
dimensions if I want it to be a rectangular area.
Also keep in mind that the person who solves it first will get
1 goat and 1 cow as a reward.
Come back to me with the answer within two days.
Every disciple set out on his
quest to look for the answer,
all of them wanted to be the
first to solve it,
not for the reward but to be in
the good books of their guru.
The majority of disciples proceeded in the following
(The following data will be shown by us as being written
by the students on their slates/notebooks)
Let x be the length and y be the breadth of the
Then 2(x+y)=26 ---------------------------------- (1)
Also it is mentioned that a goat requires 1 sq.unit
Hence 21 goats require 21 sq.units area.
Further a cow requires 2sq.units area hence 12 cows
need 24 sq.units area.
Therefore total area of shelter=
x*y = (1*21) + (2*12) = 45 sq.units --------------- (2)
From eq. (1) y = 13-x
Eq.(2) modifies to give x * (13-x) = 45
Or, 13x - x^2 = 45
Which gives x^2 - 13x + 45 = 0 ----------------------- (3)
Having just learnt quadratic
equations from their guru, they
tried to solve it by splitting the
But the more they tried the more
they became irritated and
The disciples got so involved in
this problem that they never
realized when two days just flew
The successful disciple then explained his solution as
Taking the number of cows as 12 and goats as 21
we proceed, but, according to the guru the
successful disciple would get 1 goat and 1 cow.
So instead of the original numbers, take number
of cows as 11 and goats as 20! Assuming 1 goat
and 1 cow is what goes for reward.
This is how he proceeded with his solution:
11 cows => 22 sq.units area
20 goats => 20 sq.units area
Therefore total area = 20 + 22 = 42 sq.units
x (13-x) = 42
13x – x2 = 42
x2 -13x + 42= 0
Now he just split the middle term as follows: x2 – 7x - 6x + 42=0
Which simplifies to give: (x – 6)(x - 7) = 0
Giving two solutions: x = 6 => y=7
And x = 7 => y=6
Thus the required dimensions are length = 7 units
Breadth = 6 units
Following this we give information about quadratic
equations and discriminant along with a simple graph of
a quadratic polynomial (parabola).
Finally we explain why we did not get solutions for eq (3)
that is x^2 – 13x + 45 = 0
Here the discriminant i.e. (13)^2 – 4*1*45 = (-11) which is
–ve and hence no real solution is possible.
<0 No Real Zeroes
Discriminant= =0 One Zero (two equal zeroes)
>0 Two Distinct Zeroes
The graph of a quadratic polynomial is a
The direction of opening of parabola
depends on the sign of coefficient of x2.
The vertex of the graph is given by,
(-b/2a , -D/4a)
Where D is the discriminant=b2-4ac
The direction in which the parabola opens, is called the concavity of the
The points at which the graph cuts the x-axis,( y = 0 ) are called the solutions of