QUADRATIC EQUATIONS

Directions (Ex. 1-2): In each question one or more equation(s)

is (are) provided. On the basis of these you have to find out
the relation between p and q.
Give answer (1) if p = q
Give answer (2) if p > q
Give answer (3) if q > p
Give answer (4) if p q and
Give answer (5) if q p
5 9
15 13
1. p q
28 8
14 16

K KUNDAN
2.(i) p 7 = 0
(ii) 3q 2 10q 7 0
3.(i) 4p 2 16
(ii) q 2 10q 25 0
4.(i) 4p 2 5p 1 0
(ii) q 2 2q 1 0
5.(i) q 2 11q 30 0
(ii) 2p 2 7 p 6 0

K KUNDAN

Solutions:
p 15 13 8 28 13
5 9
15 13
p q
1. 2;
or, q 14 16 9 5 3 .... (*)
28 8
14 16
p>q
Answer = (2)
Note: (*) shows that if p = 13 then q is 3.
2. 2; (i) p 7 = 0
(ii) 3q 2 10q 7 0
(i) p 7
(ii) 3p 2 3p 7q 7 0 3q (q 1) 7( q 1) 0
7
(3q 7) (q 1) 0 q or 1
3
p q Answer (2)

K KUNDAN

Note: In a quadratic equation ax 2 bx c 0

b b 2 4ac
2a
(10) (10) 2 4 3 7 10 4
7

1,
q
23
6
3
3. 3; (i) 4p 2 16
(ii) q 2 10q 25 0
x

(i) p 2

(ii) q

We see that q > p

4. 5; (i) 4p 2 5p 1 0 (ii) q 2 2q 1 0

10 100 4 1 25
5
2

5 25 16 5 3 1

,1
8
8
4
2 44
1 We see that
(ii) p
2
11 121 120 11 1

5, 6
5. 3; (i) q
2
2
7 49 48 7 1 6

,2
(ii) p
4
4
4
We see that p < q or q > p

(i) p

K KUNDAN
p q or q p

Useful points to remember about the above types of question:

In such questions three combinations of equations can be asked:
(a) Both equations are linear
(b) One equation is linear and the other quadratic
(c) Both equations are quadratic

(a) Both equations are linear

There are different methods to solve two linear equations.
Method I:
"Find the value of p in terms of q from any of the two equations and put it
in the other equation to get the value of q."
Take an example :

(i) 2p + 3q + 4 = 0
(ii)

3
5
p q 13 0
4
2

K KUNDAN
(i) p

4 3q
3
2 q .... (*)
2
2

Put it in (ii)

3
3 5
2 q q 13
4
2 2

3 9
5
q q 13
2 8
2

29
29
q
8
2

q = -4
Again we put q = -4 in (*) and get p = 4.

Thus p > q
Method II:

K KUNDAN
"Eliminate one of the two variables (p or q) by equating their
coefficients."

Take the above example:

(i) 2p + 3q + 4 = 0
(ii)

3
5
p q 13 0
4
2

(i)

5
9
+ (ii) 3 5p p 10 39 0
2
4

29p
29 p 4
4

Now, put p = 4 in (i) and get q = -4.

Thus p > q
Both the above methods are well-known to you. Adopt whichever you find
easier.
Method III: Graph method: it is of no use to us.
Method IV: Suppose the two equations are

K KUNDAN
a 1x b1 y c1 0 ... (i)

a 2 x b 2 y c 2 0 ... (ii)

If we perform (i) b2 - (ii) b1

Then

b1c 2 b 2c1
a 1b 2 a 2 b1

a 1c 2 a 2 c1 a 2 c1 a 1c 2

b1a 2 b 2 a 1 a 1b 2 a 2 b1

Note: We see that the denominators of x and y are the same. This does not
imply that x > y if b1c 2 b 2c1 a 2 c1 a 1c 2 as it fails when a 1b 2 a 2 b1 is
negative.
(b) One equation is linear and the other is quadratic:
Take Ex 2: (i) p - 7 = 0

K KUNDAN
(ii) 3q2 - 10q + 7 = 0

Equation (ii) gives two values of q. According to the given choices, both the
values of q should be either more than or less than the value of p. Why?
Because, if one value is more and the other is less than p, none of the given
choices match our answer.
Now, if both the values of q are more than p then the sum of the two values
of q should be more than 2p. And if both the values of q are less than p then
the sum of the two values of q should be less than 2p.
In the above case;
(i) p = 7

(ii) sum of two roots of q =

(10) 10

3
3

10
2 7 p > q
3
Note: When (i) p + q = 7; and
As

(ii) q2 - q - 6 = 0
In such a case, solve equation (ii). For each value of q find the corresponding
values of p from (i).

K KUNDAN

Here (ii) q = -2, 3

(i) when q = -2, p = 9 and when q = 3, p = 4 p > q
(c) Both equations are quadratic:

Ex 3, Ex 4 and Ex 5 are the examples of such questions. You can see the
most common method to solve them as given under their solutions. The
other method to solve the quadratic equations is factorisation method, which
must be known to you.
Useful conclusions:
In such cases, we can't reach to answer when one value of p is less than q
and the other value of p is more than q (the reason is the same as discussed
in (b)). So, both the values of p are either more or less than both the values
of q. This further emplies that if p1, p2, q1 and q2 are the values of p and q
then
either

p1 p 2 q1 q 2

K KUNDAN
or

p1 p 2 q1 q 2

Take Ex 5: (i) q 2 11q 30 0

(ii) 2p 2 7 p 6 0

(11)
11
1
(7) 7

(ii) Sum of roots (ie p1 + p2) =

2
2
Thus we may conclude that q > p
(i) Sum of roots (ie q1 + q2) =

But what will happen when one value of p is equal to one value of q?
For example:

2
I. p p 6 0

I. Sum of roots =

2
II. q 6q 8 0

1
1
1

II. Sum of roots =

(6)
6
1

From the above result we conclude that q > p.

But our answer is not perfect because one of the roots in the two equations
are common and our answer should be q p. Now, the problem is how can
we confirm the case of equality without getting the roots?

K KUNDAN

[I. p = 2, -3 and II q = 2, 4]
If two quadratic equations
ax 2 bx c 0 and

a 1x 2 b1x c1 0 have one common root, then

bc1 b1c ab1 a 1b

2
ac1 a 1c and vice-versa.

In the above example:

p 2 p 6 0 and
q 2 6q 8 0

K KUNDAN
1 8 6 6 1 6 (1)(1) 1 8 (1)(6)2

(8 - 36) (-7) = (14)2

196 = 196, which emplies that one root is common and hence
equality holds. So, our correct answer is q p .
The above method of checking the equality is not much time-saving.
Sometimes it is easier to get the roots.
Another Method to check equality
Suppose the common root is x. Then
I x2 x 6 0

II x 2 6 x 8 0
Now, I II gives
7x - 14 = 0 x = 2
x = 2 is the common root of the two equations.

If we put p = 2 or q = 2 in the respective equations, those should be

satisfied.
If we perform I II and get the value of p or q which satisfies the given
equation then equality must hold.

K KUNDAN

For example: I. p 2 p 6 0

II. q 2 6q 8 0

or p 2 6p 8 0 (changing q to p)
Now I II +7p - 14 = 0 p = 2
We put p = 2 in I or II. The equations hold true, which confirms that 2 is the
common root of the two equations.
Another example: I. 3p 2 7p 2 0
II. 15q 2 8q 1 0

15q 2 35q 10 0

(Put p = q in I 5)

27q 9 0

1
3

K KUNDAN

1
in I and II. As it satisfies the equations the equality holds.
3
(7) 7

Note: For final answer, Sum of roots in I =

3
3
Now, put

Sum of roots in II =

(8) 8

15
15

Therefore, our correct answer is p q .

Now let us take some more examples from Previous years papers.
Ex: (1) I. 4q 2 8q 4q 8

II. p 2 9p 2p 12

(2) I. 2p 2 40 18p
2
(3) I. 6q

1 7
q
2 2

II. q 2 13q 42
II. 12p 2 2 10p

(4) I. 4p 2 5p 1 0

II. q 2 2q 1 0

(5) I. q 2 11q 30 0

II. 2p 2 7 p 6 0

4p 8
0
5 15

K KUNDAN
(6) I.

II. 9q 2 12q 4

(7) I. q 2 15q 56 0

II. 2p 2 10p 12 0

(8) I. 18p 2 3p 3

II. 14q 2 9q 1 0

(9) I. p 2 12p 36 0

II. q 2 48 14q

(10) I. 2p 2 12p 16 0

II. 2q2 14q 24 0

(11) I. 2p 2 48 20p

II. 2q 2 18 12q

(12) I. q 2 q 2

II. p 2 7p 10 0

(13) I. p 2 36 12p

II. 4q 2 144 48q

(14) I. p 2 6p 7

II. 2q 2 13q 15 0

(15) I. 3p 2 7p 2 0

II. 2q 2 11q 15 0

(16) I. 10p 2 7p 1 0

II. 35q 2 12q 1 0

(17) I. 4p 2 25

II. 2q 2 13q 21 0

(18) I. 3p 2 7p 6

II. 6( 2q 2 1) 17q

(19) I. p 2 4

II. q 2 4q 4

(20) I. p 2 p 56

II. q 2 17q 72 0

K KUNDAN

(21) I. 3p 2 17p 10 0

II. 10q 2 9q 2 0

(22) I. p 2 3p 2 0

II. 2q 2 5q

(23) I. 2p 2 5p 2 0

II. 4q 2 1

(24) I. p 2 2p 8 0

II. q 2 2 7

(25) I. 2p 2 20p 50 0

II. q 2 25

K KUNDAN

Solution: (You are suggested to go through the detailed discussion under

the given solutions.)
(1) I 4q 2 4q 8 0

q 2 q 2 0; Sum of roots =

1
1
1

(7)
7
1
Therefore, our first conclusion is q > p. Now, check the equality:
II p 2 7 p 12 0; Sum of roots =

{1 12 - 7) (-2)} {1 7 - 1 1} = {1 12 - 1(-2)}2
or, {26} {6} = {14}2 which is not true.
Hence, our answer is q > p.
Apply another method to check the equality.
I

q2 q 2 0

II

q 2 7q 12 0

(Put p = q in II)

K KUNDAN

Apply I-II:

-6q - 14 = 0

7
3
Put this value in I or II. If we put it in I,

q=

7
49 7
49 21 18
7
20
0

20
3
9 3
9
3

10
0 which is not true. Hence our assumption that p = q is wrong.
9
Note: Such type of equation can be solved easily if we find the roots by the
method of factorisation. For example:

I q2 q 2 0

(q + 2) (q - 1) = 0 q = -2, 1

II p 2 7p 12 0 (p + 3) (p + 4) = 0 p = -3, -4
So, first try to find out the factors. If it seems difficult to factorise the
equations only then go for the other methods. The above method can be a
short cut like:
STEP 1: Multiply the coefficient of q 2 with the constant (the c in
ax 2 bx c 0 ). Here, in I, coefficient of q 2 is +1 and the constant is -2;
so the product is (+1) (-2) = -2. Now, break the coefficient of q (ie +1) in
two parts so that its product becomes -2. In this case +1 = +2, -1 are two
parts.

K KUNDAN

STEP 2: Now divide these two parts by the coefficient of q 2 , ie (+1). So

the two parts remain (+2) and (-1).
STEP 3: Now change the sign, ie +2 becomes (-2) and (-1) becomes (+1).
These are the two values or roots of the equation. See the picturised
presentation of the above method:
q2 q 2 0

S1: +2

-1

K KUNDAN

S2:

2
1

S3: -2

1
1
+1

See the solution for

II.

p 2 7p 12 0

S1:
+3
S2:

3
1

+4

4
1

S3: -3

-4

(2) I

2p 2 40 18p

II q 2 13q 42

p 2 9p 20 0
q 2 13q 42 0

Which of the three methods gives the answer easily? Naturally, the method
of factorisation. If we factorise,
(I) (p - 4) (p - 5) = 0 p = 4, 5
(II) (q - 7) (q - 6) = 0 q = 6, 7

K KUNDAN

So, answer is q > p.

See the solution by picturised presentation

(I)
p2 9p + 20 = 0
S1:

S3:

4
1
+4

5
1
+5

(II)

q 13q + 42 = 0

S1:

-7

-6

S2:

-7
1

-6
1
+6

S2:

K KUNDAN

S3: +7

See the other method (Method of assumption).

(9)
9
I sum of roots =
1
(13)
13
II sum of roots =
1
So q > p. But without checking the equality we can't confirm our answer.
So, suppose p = q. Then

p 2 9p 20 0
p 2 13p 42 0

4p 22

11
2

11
11
11
Put p =
in (I). As 9 20 0, our assumption that p = q is
2
2
2
wrong.
Therefore the final answer remains the same as q > p.

K KUNDAN

2
(3) I 6q

1 7
q
2 2

12q 2 7q 1 0

II 12p 2 2 10p 6p 2 5p 1 0
By factorisation Method:

1 1
I (3q - 1) (41 - 1) = 0 q ,
3 4
1 1
II (3p - 1) (2p - 1) = 0 p ,
3 2
So, the answer is p q .
See the solution by picturised presentation :

K KUNDAN

(I)

7
1
6q 2 q 0
2
2

S1:

-3

-3
12
S3: + 1
4
S2:

-4
-4
12
+ 13

(II)
S1:

-3

-2

-3
6
S3: + 1
2

-2
6
+ 13

S2:

K KUNDAN

By Method of Assumption:
(I) sum of roots =

(7) 7

12
12

(II) sum of roots =

(5) 5

6
6

So, p > q. But to check equality, suppose p = q. Then

12q 2 7q 1 0
6q 2 5q 1 0
Now perform (I) - 2 (II), which gives
3q - 1 = 0 q

1
3

K KUNDAN

Putting q =

1 7
1
in (I), we have 12 1 0
3
9 3

Which is true. Hence our final answer is p q.

(4) I. 4p 2 5p 1 0
II. q 2 2q 1 0

1
By Factorisation: I. (4p - 1) (p - 1) = 0 p , 1
4

II. (q - 1) (q - 1) = 0 q = 1
So, answer is q p .
Picturised presentation:
(I)

4 p 2 5p 1 0

(I)

q 2 2q 1 0

S1: -4

-1

S1: -1

-1

4
4

1
4

S2:

1
1

1
1

S2:

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1
4
By Assumption:

S3: +1

(I) sum of roots =

S3: +1

5
4

+1

(II) sum of roots = 2

Therefore q > p. Now,

Suppose p = q then
I

4 p 2 5p 1 0

4 II 4p 2 8p 4 0
3p 3 0

p 1

Put p = 1 in I. 4 - 5 + 1 = 0, which is true, hence our final answer is q p .

(5) By Factorisation
(I) (q - 6) (q - 5) = 0 q = 5, 6
3
(II) (2p - 3) (p - 2) = 0 p , 2
2
So, the answer is q > p.
Note: Try to solve these equations by picturised presentation. This saves
time as well as space for writing. Don't write Step 1, Step 2, Step 3 in three
separate lines. Change the appropriate forms in the same line to save your
time and space.
From the sum of roots it is clear that
(11)
(7)

; hence q > p.
1
2
But also suppose p = q. Now,

K KUNDAN

2p 2 7 p 6 0
2p 2 22p 60 0
15p 54 0

54 18

15 5

18
18
Putting it in I, we get 11 30 0
5
5

or, 324 - 990 + 750 0

Hence our assumption (p = q) is wrong. So, the final answer is q > p.

K KUNDAN

(6) I.

4p 8
0
5 15

12p 8 p

II. 9q 2 12q 4 0

2
3

3q 2 0 q

2
3

Therefore p = q.
(7) I q 2 15q 56 0
II p 2 5p 6 0
By Factorisation:
(I) (q - 7) (q - 8) = 0 q = 7, 8
(II) (p - 3) (p - 2) = 0 p = 2, 3
Therefore, the answer is q > p.
Note: Try to solve these two equations in a single-line step.

K KUNDAN

(8) (I) 6p 2 p 1 0
By Factorisation:

(II) 14q 2 9q 1 0

1
1
(I) (3p - 1) (2p + 1) = 0 p ,
3
2
1
1
,
(II) (7q + 1) (2q + 1) = 0 q
7
2
Therefore, the answer is p q
Note: (I)

6p 2 p 1 0

S1: +3

-2

(II) 14q 2 9q 1 0
S1:

+7

+2

S2:

3
6

S3:

1
2

2
6

S2:

1
3

S3:

7
14

2
14

1
2

1
7

1 9

, p q.
6
14
Now suppose, p = q. Then
By Assumption: As

6p 2 p 1 0

K KUNDAN
7

14p 9p 1 0 3

20p 10 0

1
2

1 1
Put it in I, Then. 6 1 0
2
2

or,

3 1
1 0 , which is true. Hence our final answer is p q .
2 2

(9) By Factorisation:
(I) p 2 12p 36 0

p 6 2 0

p=6

(II) q 2 14q 48 0 (q - 6) (q - 8) = 0 q = 6, 8
Therefore, our answer is q p.
By Assumption:

K KUNDAN
(I) Sum of roots =

( 12)
12
1

Mark that p has two values

each equal to 6

(14)
14
1
Thus q > p. Now suppose p = q. Then
(I) - (II) gives 2p - 12 = 0 or p = 6. When we put it in (I)
36 - 12 6 + 36 = 0. Which is true. Hence, the final answer is q p .
(II) Sum of roots =

(10)

(I) p 2 6p 8 0

(II) q 2 7q 12 0
By Factorisation:
(I) (p + 4) (p + 2) = 0 p = -2, -4
(II) (q + 4) (q + 3) = 0 q = -3, -4
We can't make any conclusion in such question. If we say p q, then -4
should be more than -3. Which is not true. Also, when we say q p, then
-4 should be greater than -2, which is not true. Hence we can't answer this
question. Note that although this question has been asked in a bank exam.
You are suggested to leave such questions.

K KUNDAN

(11) By Factorisation: (I) p 2 10p 24 0 (p 6) (p 4) = 0

p = 4, 6

(II) q 2 6q 9 0 (q 3) (q 3) = 0
q=3
Therefore our answer is p > q.
By Assumption: Compare the sum of roots. As 10 > 6, p > q.
15
Now, suppose, p = q and perform (I) - (II) then -4p + 15 = 0 p
.
4
2
15
15
Put it in I: 10 24 = 225 - 600 + 384 0.
4
4
Hence our final answer remains the same as p > q.
(12) By Factorisation:

K KUNDAN

(I) q 2 q 2 0 (q + 2) (q - 1) = 0 q = 1, -2
(II) p2 + 7p + 10 = 0 (p + 5) (p + 2) = 0 p = -2, -5
Therefore, the answer is q p
By Assumption: As -1 > -7, q > p
Now, put p = q and do (I) - (II) then -6p - 12 = 0 p = 2. As it satisfies
equation (I) our assumption (p = q) is true. Hence final answer is q p.
(13) (I) p 2 12p 36 0
As both are the same equations, p = q

(II) q 2 12q 36 0

(14) By Factorisation: (I) (p - 7) (p + 1) = 0 p = 7, 1

3
(II) (2q + 3) (q + 5) = 0 q , 5 .
2
Therefore, p > q
By Assumption: We compare the sum of roots.
13
6
p q.
2
Now, suppose p = q. Then II - 2 I
2p 2 13p 15 0

K KUNDAN
2p 2 12p 14 0

29
25
29
, so our assumption
None of the equations is satisfied with the value
25
(p = q) is wrong.
Note: Now onwards, the solutions by factorisation will be presented in the
picturised form; solution by assumption will not be given. You are suggested
to solve the following questions by that method also.
25p 29

(15) (I)

3p 2 7p 2 0

(II)

2q 2 11q 15 0

Step 1:

-6
6
3

Step 1:

-6
6
2

Step 2:

-1
1
3
1

Step 2:

-5
5
2
5

K KUNDAN

Step 3: +2

Step 3:

+3

(16) (I) 10p 2 7p 1 0

(II)

35q 2 12q 1 0

Step 1:

Step 1:

Therefore, q > p.

-5
-2
5
2
Step 2:
10
10
1
1

Step 3: +
2
5
Therefore, p q.

Step 2:
Step 3:

-7
7
35
1

-5
5
35
1

(17)

25
5

4
2
2
2q 13q 21 0

2
(I) 4p 25 p

(II)
Step 1:

-7
7
Step 2:
2
7
Step 3:
2

-6
6
2

K KUNDAN
+3

Therefore q > p.

(18) (I)

3p 2 7p 6 0

(II) 12q 2 17q 6 0

Step 1:

+9
9
3

Step 1:

Step 2:

Step 3: -3
Therefore, q p.
(19) (I) p2 = 4

-2
2
3
2

Step 2:
Step 3:

p = +2, -2

(II)

-9
9
12
3

-8
8
12
2

q 2 4q 4 0

Step 1: +2

2
1

+2

2
1

K KUNDAN
Step 2:

Step 3: -2

-2

Therefore, p q.
(20) (I)

p 2 p 56 0

(II)

q 2 17q 72 0

Step 1:

+8

-7

Step 1:

-8

-9

Step 2:

8
1

7
1

Step 2:

8
1

9
1

+7

Step 3:

+8

+9

Step 3: -8
Therefore, q > p.

K KUNDAN

(21) (I)

3p 2 17p 10 0

(II)

10q 2 9q 2 0

Step 1:

+15

+2

Step 1:

+5

+4

Step 2:

15
3

2
3

Step 2:

5
10

4
10

2
3

Step 3:

Step 3: -5

1
2

2
5

Therefore, q > p.
p 2 3p 2 0

(22) (I)
Step 1:

+2
2
1

+1
1
1

K KUNDAN

Step 2:

Step 3: -2

-1

(II) 2q 2 5q 0

q(2q - 5) = 0

q = 0,

Therefore, q > p.
(23) (I)

2 p 2 5p 2 0

5
2

Step 1:

+4
4
2

+1
1

Step 2:
2
1

Step 3: -2
2
1
2
2
(II) 4q 1 q
4

1
1
q ,
2
2

K KUNDAN

Therefore, q p
(24) (I)

p 2 2p 8 0

Step 1:

+4

4
1
Step 3: -4
(II) q2 = 9
Step 2:

-2

2
1
+2
q = 3, 3

Although this question is from the previous paper asked in BSRB Mumbai,
yet no conclusions can be drawn. You are suggested to leave such questions.
(25) (I)

p 2 10p 25 0

K KUNDAN

Step 1:

+5
5
Step 2:
1
Step 3: -5
(II) q 2 25

+5
5
1
-5

q 5, 5

Therefore, q p.