1. In This presentation we deal with the solving of Quadratic Equations by splitting the middle term..

2. 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. 3. 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 students: 4. 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. NOW GO!!! 5. 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. 6. The majority of disciples proceeded in the followingmanner: (The following data will be shown by us as being writtenby the students on their slates/notebooks)Let x be the length and y be the breadth of the shelter. Then2(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 cowsneed 24 sq.units area.Therefore total area of shelter= x*y = (1*21) + (2*12) = 45 sq.units --------------- (2) From eq. (1) y = 13-xEq.(2) modifies to give x * (13-x) = 45Or,13x - x^2 = 45 Which gives x^2 - 13x + 45 = 0 ----------------------- (3) 7. Having just learnt quadraticequations from their guru, they tried to solve it by splitting the middle term. But the more they tried the morethey became irritated and exhausted.The disciples got so involved inthis problem that they never realized when two days just flew past. 8. The successful disciple then explained his solution asfollows:Taking the number of cows as 12 and goats as 21we proceed, but, according to the guru thesuccessful disciple would get 1 goat and 1 cow.So instead of the original numbers, take numberof cows as 11 and goats as 20! Assuming 1 goat and 1 cow is what goes for reward. 9. 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 units7 units6 units 10. 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 Two Distinct Zeroes 11. The graph of a quadratic polynomial is a parabola. 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 12. a>0a