Quadratic Equations for Bank Po

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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 outthe 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 qp and Give answer (5) if pq

1.5 9 15 13p q28 8 14 16

2.(i) p – 7 = 0 (ii) 07q10–q3 2

3.(i) 16p4 2 (ii) 025q10–q2

4.(i) 01p5–p4 2 (ii) 01q2–q2

5.(i) 030q11–q2 (ii) 06p7–p2 2

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Solutions:

1. 2; 5 9 15 13p q28 8 14 16

or, p 15 13 8 28 13 .... (*)q 14 16 9 5 3

p > q Answer = (2)Note: (*) shows that if p = 13 then q is 3.2. 2; (i) p – 7 = 0 (ii) 07q10q3 2

(i) 7p (ii) 0)1–q(7–)1–q(q307q7p3p3 2

1or37q0)1–q()7–q3(

)2(Answerqp

Note: In a quadratic equation 0cbxax 2

a2ac4bbx

2

37,1

6410

32734–)10(–)10(––

q2

3. 3; (i) 16p4 2 (ii) 025q10–q2

(i) 2p (ii) 52

2514–10010q

We see that q > p4. 5; (i) 01p5–p4 2 (ii) 01q2–q2

(i) 1,41

835

816–255p

(ii) 12

4–42p

We see that p q or q p

5. 3; (i) 6,52

1112

120–12111q

(ii) 2,46

417

448–497p

We see that p < q or q > pUseful 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

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(a) Both equations are linearThere 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 itin the other equation to get the value of q."Take an example : (i) 2p + 3q + 4 = 0

(ii) 013q25p

43

(i) q232

2q34p

.... (*)

Put it in (ii) 13q25q

232

43

13q25q

89

23

229q

829

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

Method II:"Eliminate one of the two variables (p or q) by equating theircoefficients."

Take the above example: (i) 2p + 3q + 4 = 0

(ii) 013q25p

43

(i) 25

+ (ii) ×3 03910p49p5

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4p294p29

Now, put p = 4 in (i) and get q = -4.Thus p > qBoth the above methods are well-known to you. Adopt whichever you findeasier.Method III: Graph method: it is of no use to us.Method IV: Suppose the two equations are

0cybxa 111 ... (i)

0cybxa 222 ... (ii)

If we perform (i) ×b2 - (ii) ×b1

Then1221

1221

babacbcbx

1221

2112

1221

1221

babacaca

ababcacay

Note: We see that the denominators of x and y are the same. This does not

imply that x > y if 21121221 cacacbcb as it fails when 1221 baba isnegative.(b) One equation is linear and the other is quadratic:Take Ex 2: (i) p - 7 = 0

(ii) 3q2 - 10q + 7 = 0Equation (ii) gives two values of q. According to the given choices, both thevalues 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 givenchoices match our answer.Now, if both the values of q are more than p then the sum of the two valuesof q should be more than 2p. And if both the values of q are less than p thenthe sum of the two values of q should be less than 2p.In the above case;(i) p = 7

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(ii) sum of two roots of q = 310

3)10(

As 723

10 p > q

Note: When (i) p + q = 7; and (ii) q2 - q - 6 = 0

In such a case, solve equation (ii). For each value of q find the correspondingvalues of p from (i).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 themost common method to solve them as given under their solutions. Theother method to solve the quadratic equations is factorisation method, whichmust be known to you.Useful conclusions:In such cases, we can't reach to answer when one value of p is less than qand the other value of p is more than q (the reason is the same as discussedin (b)). So, both the values of p are either more or less than both the valuesof q. This further emplies that if p1, p2, q1 and q2 are the values of p and qthen

either 2121 qqpp

or 2121 qqpp

Take Ex 5: (i) 030q11q2

(ii) 06p7p2 2

(i) Sum of roots (ie q1 + q2) = 111

)11(

(ii) Sum of roots (ie p1 + p2) = 27

2)7(

Thus we may conclude that q > pBut what will happen when one value of p is equal to one value of q?For example:

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I. 06pp2 II. 08q6q2

I. Sum of roots = 111

II. Sum of roots = 61

)6(

From the above result we conclude that q > p.But our answer is not perfect because one of the roots in the two equationsare common and our answer should be q p. Now, the problem is how canwe confirm the case of equality without getting the roots?[I. p = 2, -3 and II q = 2, 4]

If two quadratic equations

0cbxax 2 and

0cxbxa 112

1 have one common root, then

baabcbbc 1111 211 caac and vice-versa.

In the above example:

06pp2 and

08q6q2

2)6)(1(81)1)(1(616681

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

196 = 196, which emplies that one root is common and henceequality holds. So, our correct answer is pq .

The above method of checking the equality is not much time-saving.Sometimes it is easier to get the roots.Another Method to check equalitySuppose the common root is x. Then

I 06xx 2

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II 08x6x 2

Now, I – II gives7x - 14 = 0 x = 2x = 2 is the common root of the two equations. If we put p = 2 or q = 2 in the respective equations, those should besatisfied. If we perform I – II and get the value of p or q which satisfies the givenequation then equality must hold.

For example: I. 06pp2

II. 08q6q2

or 08p6p2 (changing q to p)

Now I – II +7p - 14 = 0 p = 2We put p = 2 in I or II. The equations hold true, which confirms that 2 is thecommon root of the two equations.

Another example: I. 02p7p3 2

II. 01q8q15 2

(Put p = q in I × 5)

09q27

010q35q15 2

31q

Now, put 31

in I and II. As it satisfies the equations the equality holds.

Note: For final answer, Sum of roots in I = 37

3)7(

Sum of roots in II = 158

15)8(

Therefore, our correct answer is qp .

Now let us take some more examples from Previous years’ papers.

Ex: (1) I. 8q4q8q4 2 II. 12p2p9p2

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(2) I. p1840p2 2 II. 42q13q2

(3) I. q27

21q6 2 II. p102p12 2

(4) I. 01p5p4 2 II. 01q2q2

(5) I. 030q11q2 II. 06p7p2 2

(6) I. 0158

5p4

II. 4q12q9 2

(7) I. 056q15q2 II. 012p10p2 2

(8) I. 3p3p18 2 II. 01q9q14 2

(9) I. 036p12p2 II. q1448q2

(10) I. 016p12p2 2 II. q q 22 14 24 0

(11) I. p2048p2 2 II. q1218q2 2

(12) I. 2qq2 II. 010p7p2

(13) I. p1236p2 II. q48144q4 2

(14) I. 7p6p2 II. 015q13q2 2

(15) I. 02p7p3 2 II. 015q11q2 2

(16) I. 01p7p10 2 II. 01q12q35 2

(17) I. 25p4 2 II. 021q13q2 2

(18) I. 6p7p3 2 II. q17)1q2(6 2

(19) I. 4p2 II. 4q4q2

(20) I. 56pp2 II. 072q17q2

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(21) I. 010p17p3 2 II. 02q9q10 2

(22) I. 02p3p2 II. q5q2 2

(23) I. 02p5p2 2 II. 1q4 2

(24) I. 08p2p2 II. 72q2

(25) I. 050p20p2 2 II. 25q2

Solution: (You are suggested to go through the detailed discussion underthe given solutions.)

(1) I 08q4q4 2 ;02qq2 Sum of roots = 111

II ;012p7p2 Sum of roots = 71

)7(

Therefore, our first conclusion is q > p. Now, check the equality:{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 02qq2

II 012q7q2 (Put p = q in II)Apply I-II: -6q - 14 = 0

q = 37

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

0237

37 2

0237

949

09

182149

09

10 which is not true. Hence our assumption that p = q is wrong.

Note: Such type of equation can be solved easily if we find the roots by themethod of factorisation. For example:

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I 02qq2 (q + 2) (q - 1) = 0 q = -2, 1

II 012p7p2 (p + 3) (p + 4) = 0 p = -3, -4So, first try to find out the factors. If it seems difficult to factorise theequations only then go for the other methods. The above method can be ashort cut like:

STEP 1: Multiply the coefficient of 2q with the constant (the c in

0cbxax 2 ). Here, in I, coefficient of 2q is +1 and the constant is -2;so the product is (+1) (-2) = -2. Now, break the coefficient of q (ie +1) intwo parts so that its product becomes -2. In this case +1 = +2, -1 are twoparts.

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

02qq2

S1: +2 -1

S2: 12

11

S3: -2 +1See the solution for

II. 012p7p2

S1: +3 +4

S2: 13

14

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S3: -3 -4

(2) I p1840p2 2 020p9p2

II 42q13q2 042q13q2

Which of the three methods gives the answer easily? Naturally, the methodof factorisation. If we factorise,(I) (p - 4) (p - 5) = 0 p = 4, 5(II) (q - 7) (q - 6) = 0 q = 6, 7So, answer is q > p.See the solution by picturised presentation(I) p2 – 9p + 20 = 0

S1: –4 –5

S2: 14

15

S3: +4 +5

S1:

S2:

S3:

(II) q – 13q + 42 = 02

-7 -6

-7 -61 1

+7 +6

See the other method (Method of assumption).

I sum of roots = 91

)9(

II sum of roots = 131

)13(

So q > p. But without checking the equality we can't confirm our answer.So, suppose p = q. Then

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22p4

042p13p

020p9p2

2

211p

Put p = 211

in (I). As ,0202

1192

11 2

our assumption that p = q is

wrong.Therefore the final answer remains the same as q > p.

(3) I q27

21q6 2 01q7q12 2

II p102p12 2 01p5p6 2

By factorisation Method:

I (3q - 1) (41 - 1) = 0 41,

31q

II (3p - 1) (2p - 1) = 0 21,

31p

So, the answer is qp .

See the solution by picturised presentation :

(I) 021q

27q6 2

S1:

S2:

S3:

-3 -4

-3 -412 12 1 14 3+ +

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S1:

S2:

S3:

-3 -2

-3 -26 6 1 12 3+

(II)

+

By Method of Assumption:

(I) sum of roots = 127

12)7(

(II) sum of roots = 65

6)5(

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

01q7q12 2

01q5q6 2

Now perform (I) - 2 × (II), which gives

3q - 1 = 0 31q

Putting q = 31

in (I), we have 0137

9112

Which is true. Hence our final answer is .qp

(4) I. 01p5p4 2

II. 01q2q2

By Factorisation: I. (4p - 1) (p - 1) = 0 1,41p

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

(I) 01p5p4 2 (I) 01q2q2

S1: -4 -1 S1: -1 -1

S2: 44

41

S2: 11

11

S3: +1 41

S3: +1 +1

By Assumption:

(I) sum of roots = 45

(II) sum of roots = 2Therefore q > p. Now,Suppose p = q then

1p03p3

04p8p4II4

01p5p4I2

2

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

(5) By Factorisation(I) (q - 6) (q - 5) = 0 q = 5, 6

(II) (2p - 3) (p - 2) = 0 2,23p

So, the answer is q > p.Note: Try to solve these equations by picturised presentation. This savestime as well as space for writing. Don't write Step 1, Step 2, Step 3 in threeseparate lines. Change the appropriate forms in the same line to save yourtime and space.From the sum of roots it is clear that

2)7(

1)11(

; hence q > p.

But also suppose p = q. Now,

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054p15

060p22p2

06p7p22

2

518

1554p

Putting it in I, we get 0305

18115

18 2

or, 324 - 990 + 750 0Hence our assumption (p = q) is wrong. So, the final answer is q > p.

(6) I. 0158

5p4

32p8p12

II. 04q12q9 2 32q02q3 2

Therefore p = q.

(7) I 056q15q2 II 06p5p2 By Factorisation:(I) (q - 7) (q - 8) = 0 q = 7, 8(II) (p - 3) (p - 2) = 0 p = 2, 3Therefore, the answer is q > p.Note: Try to solve these two equations in a single-line step.

(8) (I) 01pp6 2 (II) 01q9q14 2 By Factorisation:

(I) (3p - 1) (2p + 1) = 0 21,

31p

(II) (7q + 1) (2q + 1) = 0 21,

71q

Therefore, the answer is qp

Note: (I) 01pp6 2 (II) 01q9q14 2

S1: +3 -2 S1: +7 +2

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S2: 63

62

S2: 147

142

S3: 21

31

S3: 21

71

By Assumption: As qp,14

961

.Now suppose, p = q. Then

010p20

301p9p14

701pp62

2

21p

Put it in I, Then. 0121

216

2

or, 0121

23

, which is true. Hence our final answer is qp .

(9) By Factorisation:

(I) 036p12p2 06p 2 p = 6

(II) 048q14q2 (q - 6) (q - 8) = 0 q = 6, 8

Therefore, our answer is .pq

By Assumption:

(I) Sum of roots =

6toequaleach

valuestwohaspthatMark12

1)12(

(II) Sum of roots = 141

)14(

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 pq .

(10) (I) 08p6p2

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(II) 012q7q2

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 -4should 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 thisquestion. Note that although this question has been asked in a bank exam.You are suggested to leave such questions.

(11) By Factorisation: (I) 024p10p2 (p – 6) (p – 4) = 0 p = 4, 6

(II) 09q6q2 (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.

Now, suppose, p = q and perform (I) - (II) then -4p + 15 = 0 415p .

Put it in I: 244

15104

15 2

= 225 - 600 + 384 0.

Hence our final answer remains the same as p > q.

(12) By Factorisation:

(I) 02qq2 (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 pq By Assumption: As -1 > -7, q > pNow, put p = q and do (I) - (II) then -6p - 12 = 0 p = –2. As it satisfiesequation (I) our assumption (p = q) is true. Hence final answer is .pq

(13) (I) 036p12p2 (II) 036q12q2 As both are the same equations, p = q

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

(II) (2q + 3) (q + 5) = 0 5,23q .

Therefore, p > qBy Assumption: We compare the sum of roots.

.qp2136

Now, suppose p = q. Then II - 2 × I

29p25

014p12p2

015p13p22

2

2529p

None of the equations is satisfied with the value ,2529

so our assumption

(p = q) is wrong.Note: Now onwards, the solutions by factorisation will be presented in thepicturised form; solution by assumption will not be given. You are suggestedto solve the following questions by that method also.

(15) (I) 02p7p3 2 (II) 015q11q2 2

Step 1: -6 -1 Step 1: -6 -5

Step 2: 36

31

Step 2: 26

25

Step 3: +2 31

Step 3: +3 25

Therefore, q > p.

(16) (I) 01p7p10 2 (II) 01q12q35 2

Step 1: -5 -2 Step 1: -7 -5

Step 2: 105

102

Step 2: 357

355

Step 3: + 21

51

Step 3: 51

71

Therefore, p q.

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(17) (I) 25

425p25p4 2

(II) 021q13q2 2 Step 1: -7 -6

Step 2: 27

26

Step 3: 27

+3 Therefore q > p.

(18) (I) 06p7p3 2 (II) 06q17q12 2

Step 1: +9 -2 Step 1: -9 -8

Step 2: 39

32

Step 2: 129

128

Step 3: -3 32

Step 3: 43

32

Therefore, q p.

(19) (I) p2 = 4 p = +2, -2 (II) 04q4q2

Step 1: +2 +2

Step 2: 12

12

Step 3: -2 -2

Therefore, p q.

(20) (I) 056pp2 (II) 072q17q2

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Step 1: +8 -7 Step 1: -8 -9

Step 2: 18

17

Step 2: 18

19

Step 3: -8 +7 Step 3: +8 +9

Therefore, q > p.

(21) (I) 010p17p3 2 (II) 02q9q10 2

Step 1: +15 +2 Step 1: +5 +4

Step 2: 315

32

Step 2: 105

104

Step 3: -5 32

Step 3: 21

52

Therefore, q > p.

(22) (I) 02p3p2

Step 1: +2 +1

Step 2: 12

11

Step 3: -2 -1

(II) 0q5q2 2

q(2q - 5) = 0 q = 25,0

Therefore, q > p.

(23) (I) 02p5p2 2

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Step 1: +4 +1

Step 2: 24

21

Step 3: -2 21

(II) 41q1q4 22 2

1,21q

Therefore, pq

(24) (I) 08p2p2

Step 1: +4 -2

Step 2: 14

12

Step 3: -4 +2(II) q2 = 9 q = ±3, –3Although 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) 025p10p2

Step 1: +5 +5

Step 2: 15

15

Step 3: -5 -5

(II) 5,5q25q2 Therefore, q p.