1.If ‘r’ is a root of the equation,
ax  bx  c  0
2
then it satisfies the equation, that is
ar  br  c  0
2
1.
2.
Assignment
3.
2.Discriminant =D = b  4 ac
D>0, real and unequal roots
D=0, equal roots
D<0, no real roots
2
4.
5.
6.
7.
Assignment
8.
9.
10.
11.
3. Solving by factorization that is
splitting the middle term.
12.
13.
14.
Note that in questions like the two above, the splitting is the same as given in the
problem. In the second one just open the brackets and take common.
15.
Assignment
16.
Note that both the first and the last terms of the expression on the LHS are perfect
squares. That should prompt one to check whether the entire expression is a
perfect square.
The equation can also be written as :
a  b  x 2  2 ( a  b ) x 4 ( a  b )   4 ( a  b ) 2
0
4. Quadratic formula
17.
18.
5. Completion of square method
19.(i)
(ii)
The equations above can be solved easily by both quadratic formula as well as
completion of square method.
Try to solve it using both!!
6. Which method to use to solve a
quadratic equation if not mentioned in
question?
• If the coefficients are small, you should first
try to split the middle term. If the split is not
clear or is not striking quickly then move to
another method.
• Use completion of square method only if the
coefficient of x2 is 1 or is a perfect square. If
you can remember the common binomial
squares or even learn to recognise them this
method can be very handy in some cases.
• Examples of binomial squares are :
x  2 x  1   x  1
2
2
x  6 x  9  x  3
2
2
4 x  4 x  1  2 x  1
2
2
• The quadratic formula method works in all cases.
But when the split is clear the calculations in the
formula method may be unnecessarily tedious.
• Be perfect with the quadratic formula!!
7. Word problems method
• Since the word problems have rational or
integer solutions most of the time after
getting the quadratic equation , first try to
solve it by splitting the middle term.
• Do remember to discard the irrelevant
solution and write the final answer.
•
8.Framing equations for word
problems
(simpler
ones)
Consider the following problem :
20.
• Look for what is asked. Take it as x.
• In this case let the number of packets bought be ‘x’. Then 4 more packets
means x+4 packets. Now the problem reads :
• A shopkeeper buys x packets of biscuits for Rs 80. If he had bought x+4
packets for same amount (that is 80), each packet would have cost Re. 1
less.
• Thus new cost of each packet = old cost of each packet – 1
• Cost of each packet = total cost /no. of packets.
• Now you should be able to get the equation. Simplify and solve it to get
16 as your answer. You will also get –20 as your answer which you can
discard.
Now try these:
21.
22.
23.
24.
8.Framing equations for word
problems
(digit problem)
25.
In these problems, instead of taking the no. as ‘x’, take one of the digits as ‘x’.
Let the ones digit be ‘x’ in this problem, then given product of digits is 18, the tens digit
will be 18 .
x
Thus the number=
18
 10  x  1
x
18
x

10

1
And the number obtained by interchanging the digits =
x
Now frame the equation, multiply the equation by ‘x’ to remove ‘x’ from the
denominator and you will end up with a quadratic equation to solve.
• Similar to the digit problem is the fraction
problem where either numerator or
denominator is taken as ‘x’. Try these
problems:
26.
9.Framing equations for word
problems
(distance time problem)
•
Consider the following problem:
•
•
We start by taking usual speed of the train as ‘x’ km/hr.
There are two cases in the problem.First when the speed is ‘x’ and second case when the speed is
increased by 5 km/hr that is the speed is (x+5) km/hr.
The distance in both cases is 150 km.
Now we can find the time taken in both the cases.
Time = distance/speed
First case the time taken is 150/x hr
Second case the time taken is 150/(x+5) hr
Given time taken in second case is one hour less than the time taken in first case. That is
27.
•
•
•
•
•
•
150
•
x5
Now solve the equation.

150
x
1
Now try these problems:
28.
29.
30.
31.
10.Framing equations for word
problems
(work -time problem or tap problem)
• Remember in work time problems time taken
is inversely propotional to number of men or
number of taps. So if the time taken by one
man is t1 , the time taken by another man is
t2 and the time taken by both together to
finish a work is t, then t  t1  t 2
• Rather the correct relation is
1
t1

1
t2

1
t
• Consider the following problem
32.
• Let the time taken by smaller tap be x hrs.
• Then the time taken by larger tap will be (x – 10) hrs
• Thus from the previous slide
•
1
x

1
x  10

1
9
3
or
1
x

1
x  10
8
• Now simplify and solve the equation

8
75
Now try these problems:
33.
34.
35.
36.
ANSWERS