Quadratic Equation Questions For Bank Clerk Prelims

The aspirants are all searching for detailed information on Quadratic Equation for competitive exams. Our IBPS Guide team is providing new series of Quadratic Equation questions for competitive exams. Quadratic Equation is an important topic in quantitative aptitude. The quadratic equation is one of the easiest and tricky parts of the quantitative aptitude section. This section requires good observation to avoid silly mistakes during the examination. But with good practice of ample questions in the Quadratic Equation For competitive exams. The topic is very common in all competitive exams and requires a lot of practice. It is very different than the linear equation. The main difference between the quadratic equation and linear equation is that the latter will have a term x² in it. When the power of a term is 2 it is called a quadratic term or a term with a degree 2. You can expect 5 questions from it in almost all the bank exams. These questions are very easy to solve within 2 minutes if you have practiced well. You can increase your chances of selection by solving these questions before approaching other questions.

Candidates must dedicate quality time in the Quadratic Equation For bank PO Mains examination PDF practice to ace this topic. Solving questions in the numerical ability section requires good accuracy. Candidates can expect 3-5 questions in the quadratic equation in the bank PO Mains examination. Utilize the Quadratic Equation For bank PO Mains examination PDF provided and ace the concepts with the help of solutions provided with detailed explanations. Dear Aspirants, Our IBPS Guide team is providing new series of quantitative Questions for competitive exams so the aspirants can practice it on a daily basis. These questions are framed by our skilled experts after understanding your needs thoroughly. Aspirants can practice these new series of questions daily to familiarize themselves with the exact exam pattern and make their preparation effective.

Quadratic Equation for competitive exams (FAQ):

What are the 5 methods of a quadratic equation?

Ans. There are several methods you can use to solve a quadratic equation: Factoring Completing the Square Quadratic Formula Graphing Factoring, Completing the Square, Quadratic Formula, and Graphing.

2) What are 3 ways to solve quadratic equations?

Ans. when a 0. There are three basic methods for solving quadratic equations: factoring, using the quadratic formula, and completing the square

 

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Directions (01-10): Following question contains two equations as I and II. You have to solve both equations and determine the relationship between them and give an answer as,

1) I) x² + 28x + 115 = 0

II)y²+ 29y + 138 = 0

a) x > y

b) x ≥ y

c) x = y or relationship can’t be determined.

d) x < y

e) x ≤ y

2) I) 2x² – 30x + 108 = 0

II)y²– 15y + 54 = 0

a) x > y

b) x ≥ y

c) x = y or relationship can’t be determined.

d) x < y

e) x ≤ y

3) I) x² + 18x+72 =0

II) y²+ 6y +8 =0

a) x > y

b) x ≥ y

c) x = y or relationship can’t be determined.

d) x < y

e) x ≤ y

4) I) x² – 18x + 17 = 0

II)y² + 9y – 10 = 0

a) x > y

b) x ≥ y

c) x = y or relationship cannot be determined.

d) x < y

e) x ≤ y

5) I) x² – 2x – 3 = 0

II)y² + 18y + 72 = 0

a) x > y

b) x ≥ y

c) x = y or relationship cannot be determined.

d) x < y

e) x ≤ y

6) I) x² – 12x + 32 = 0

II)y²–8y + 12 = 0

a) x > y

b) x ≥ y

c) x = y or relationship cannot be determined.

d) x < y

e) x ≤ y

7) I) x² – 14x – 120 = 0

II)y²– 18y – 19 = 0

a) x > y

b) x ≥ y

c) x = y or relationship can’t be determined.

d) x < y

e) x ≤ y

8) I) 2x + y = 23

II)3x + 2y = 38

a) x > y

b) x ≥ y

c) x = y or relationship can’t be determined.

d) x < y

e) x ≤ y

9) I) x² + 189 = 310

II)y²– 12y + 11 = 0

a) x > y

b) x ≥ y

c) x = y or relationship can’t be determined.

d) x < y

e) x ≤ y

10) I) x² + 12x + 20 = 0

II)y² + 13y + 22 = 0

a) x > y

b) x ≥ y

c) x = y or relationship can’t be determined.

d) x < y

e) x ≤ y

Answers :

1) Answer: C

x² + 28x + 115 = 0

x² + 23x + 5x + 115 = 0

x(x + 23) + 5(x + 23) = 0

(x + 5) (x + 23) = 0

x = -5, -23

y² + 29y + 138 = 0

y² + 23y + 6y + 138 = 0

y(y + 23) + 6(y + 23) = 0

(y + 6)(y + 23) = 0

y = -6, -23

Relationship between x and y cannot be established.

2) Answer: C

2x² – 30x + 108 = 0

2x² – 12x – 18x + 108 = 0

2x(x – 6) – 18(x – 6) = 0

(2x – 18)(x – 6) = 0

x = 9, 6

y² – 15y + 54 = 0

y² – 9y – 6y + 54 = 0

y(y – 9) – 6(y – 9) = 0

(y – 6)(y – 9) = 0

y = 6, 9

Relationship between x and y cannot be established.

3) Answer: D

I) x²+ 18x+72 =0

x² + 12x +6x +72 =0

x(x+12)+6 (x+12) =0

(x+6) (x+12) =0

x=-6, -12

II) y²+ 6y +8 =0

y² + 4y+2y +8 =0

y(y+4)+2 (y+4) =0

(y+2) (y+4) =0

y =-2, -4

x<y

4) Answer: B

x² – 18x + 17 = 0

x² – 17x – x + 17 = 0

x(x – 17) – 1(x – 17) = 0

(x – 17) (x – 1) = 0

x = 17, 1

y² + 9y – 10 = 0

y² + 10y – y – 10 = 0

y(y + 10) – 1(y + 10) = 0

(y – 1) (y + 10) = 0

y = 1, -10

Hence, x≥y

5) Answer: A

x² – 2x – 3 = 0

x² – 3x + x – 3 = 0

x(x – 3) + 1(x – 3) = 0

(x + 1) (x – 3) = 0

x =-1, 3

y² + 18y + 72 = 0

y² + 12y + 6y + 72 = 0

y(y + 12) + 6(y + 12) = 0

(y + 6)(y + 12) = 0

y = -6, -12

Hence, x>y

6) Answer: C

x² – 12x + 32 = 0

x² – 8x – 4x + 32 = 0

x(x – 8) – 4(x – 8) = 0

(x – 4)(x – 8) = 0

x = 4, 8

y² – 8y + 12 = 0

y² – 6y – 2y + 12 = 0

y(y – 6) – 2(y – 6) = 0

(y – 2) (y – 6) = 0

y = 2, 6

Relationship between x and y cannot be established.

7) Answer: C

x² – 14x – 120 = 0

x² – 20x + 6x – 120 = 0

x(x – 20) + 6(x – 20) = 0

(x + 6)(x – 20) = 0

x = -6, 20

y² – 18y – 19 = 0

y² – 19y + y – 19 = 0

y(y – 19) + 1(y – 19) = 0

( y + 1)(y – 19) = 0

y = -1, 19

Relationship between x and y cannot be established.

8) Answer: A

2x + y = 23 ——–(1)

3x + 2y = 38 ——-(2)

(1) * 2 – (2)

x = 8

y = 23 – 16 = 7

x > y

9) Answer: C

I) x²+ 189 = 310

x² = 121

x = 11, -11

II) y²– 12y + 11 = 0

y² – 11y –y + 11 = 0

y(y – 11) – 1(y – 11) = 0

(y – 1)(y – 11) = 0

y = 1, 11

Relationship between x and y cannot be established.

10) Answer: C

x² + 12x + 20 = 0

x² + 10x + 2x + 20 = 0

x(x + 10) + 2(x + 10) = 0

(x + 2)(x + 10) = 0

x= -2, -10

y² + 13y + 22 = 0

y² + 11y + 2y + 22 = 0

y(y + 11) + 2(y + 11) = 0

(y + 2)(y + 11) = 0

y = -2, -11

Relationship between x and y cannot be established.

 

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Directions (11-20): In each of the following questions, two equations are given. You have to solve both the equations to find the relation between x and y.

11) I) x² – 18x = 144

II) y²+ 9y = 90

A.x > y

B.x ≥ y

C.x = y or relationship can’t be determined.

D.x < y

E.x ≤ y

12) I) x² – 13x + 42 = 0

II)y²– 19y + 88 = 0

A.x > y

B.x ≥ y

C.x = y or relationship can’t be determined.

D.x < y

E.x ≤ y

13) I) x² + 17x – 18 = 0

II)2y²+ 16y + 32 = 0

A.x > y

B.x ≥ y

C.x = y or relationship can’t be determined.

D.x < y

E.x ≤ y

14) I) x² – 5x + 6 = 0

II) y² – 9y + 20 = 0

A.x < y

B.x > y

C.x ≤ y

D.x ≥ y

E.x=y or the relationship between x and y cannot be determined

15) I) x² + x – 2 = 0         

II) y² -3y + 2 = 0

A.x < y

B.x > y

C.x ≤ y

D.x ≥ y

E.x=y or the relationship between x and y cannot be determined

16) I) 3x² + 30x + 27 = 0

II)y²– 28y – 29 = 0

A.x > y

B.x ≥ y

C.x = y or relationship can’t be determined.

D.x < y

E.x ≤ y

17) I) x² – x – 20 = 0

II)y²– 13y + 40 = 0

A.x > y

B.x ≥ y

C.x = y or relationship can’t be determined.

D.x < y

E.x ≤ y

18) I)  x² -18x +81=0

II) y²+ 18y +81 = 0

A.x > y

B.x ≥ y

C.x = y or relationship can’t be determined.

D.x < y

E.x ≤ y

19) I) x²-16x+48 = 0

II) y²-9y +18 = 0

A.x > y

B.x ≥ y

C.x = y or relationship can’t be determined.

D.x < y

E.x ≤ y

20) I) x² +11x +30 = 0

II)y²+9y +20 = 0

A.x > y

B.x ≥ y

C.x = y or relationship can’t be determined.

D.x < y

E.x ≤ y

Answers :

11) Answer: C

x² – 18x = 144

x² – 18x – 144 = 0

x² – 24x + 6x – 144 = 0

x(x – 24) + 6(x – 24) = 0

(x + 6)(x – 24) = 0

x = -6, 24

y² + 9y = 90

y² + 9y – 90 = 0

y² + 15y – 6y – 90 = 0

y(y + 15) – 6(y + 15) = 0

(y – 6) (y + 15) = 0

y = 6, -15

Relationship between x and y cannot be determined.

12) Answer: D

x² – 13x + 42 = 0

x² – 6x – 7x + 42 = 0

x(x – 6) – 7(x – 6) = 0

(x – 7)(x – 6) = 0

x = 7, 6

y² – 19y + 88 = 0

y² – 11y – 8y + 88 = 0

y(y – 11) – 8(y – 11) = 0

(y – 11)(y – 8) = 0

y = 11, 8

x < y

13) Answer: C

x² + 17x – 18 = 0

x² + 18x – x – 18 = 0

x(x + 18) – 1(x + 18) = 0

(x + 18)(x – 1) = 0

x = -18, 1

2y² + 16y + 32 = 0

2y² + 8y + 8y + 32 = 0

2y(y + 4) + 8(y + 4) = 0

(2y + 8)(y + 4) = 0

y = -4, -4

Relationship between x and y cannot be determined.

14) Answer: A

x² – 5x + 6 = 0

=>x² – 3x – 2x + 6 = 0

=> x(x – 2) -3(x – 2) = 0

=> (x – 2) (x – 3) = 0

=> x = 2, 3

y² – 9y + 20 = 0

y² – 5y -4y + 20 = 0

=>y (y – 5) – 4 (y – 5) = 0

=> (y – 5) (y – 4) = 0

=> y = 4, 5

Hence, x < y

15) Answer: C

x² + x – 2 = 0

x² + 2x – x – 2 = 0

=>x(x + 2) – 1 (x + 2) = 0

=>(x – 1) (x + 2) = 0

=> x = 1, -2       

y² – 3y + 2 = 0

y² – 2y -y + 2 = 0

=> y(y – 2) -1 (y – 2) = 0

=> (y – 1) (y – 2) = 0

=> y = 1, 2

Hence, x ≤ y

16) Answer: E

3x² + 30x + 27 = 0

3x² + 27x + 3x + 27 = 0

3x(x + 9) + 3(x + 9) = 0

(3x + 3)(x + 9) = 0

x = – 1, -9

y² – 28y – 29 = 0

y² – 29y + y – 29 = 0

y(y – 29) + 1(y – 29) = 0

(y + 1)(y – 29) = 0

y = -1, 29

x ≤ y

17) Answer: E

x² – x – 20 = 0

x² – 5x + 4x – 20 = 0

x(x – 5) + 4(x – 5) = 0

(x – 5)(x + 4) = 0

x = 5, -4

y² – 13y + 40 = 0

y² – 8y – 5y + 40 = 0

y (y – 8) – 5 (y – 8) = 0

(y – 5)(y – 8) = 0

y = 5, 8

y ≥ x

18) Answer:  A

x² -18x +81=0

x² -9x-9x +81=0

x(x-9) -9(x-9) = 0

x= +9, +9

y² + 18y +81 = 0

y² + 9y+9y +81 = 0

y(y+9)+9(y+9) = 0

(y+9)(y+9)= 0

y= -9, -9

Hence x > y

19) Answer: C

x²-16x+48 = 0

x²-12x-4x+48 = 0

x(x-12) -4(x-12)= 0

(x-4)(x-12) =0

x=  4, 12

y² -9y +18 = 0

y² -3y-6y +18 = 0

y(y-3) –6(y-3) = 0

(y-6)(y-3) = 0

y= +6, +3

Hence, the relationship between x and y can’t be determined.

20) Answer: E

x² +11x +30 = 0

x² +6x +5x  +30 = 0

x(x+6) +5(x+6) = 0

(x+5)(x+6) = 0

x= -5, -6

y² +9y +20 = 0

y² +4y + 5y+20 = 0

y(y+4) + 5(y+4) = 0

(y+5)(y+4) = 0

y= -5, -4

Hence, x ≤ y

 

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Directions (21-30): In each of the following questions, two equations are given. You have to solve both the equations to find the relation between x and y.

21) I: x² + 6x + 8 = 0

II: y² -8y + 15 = 0

A.x < y

B.x > y

C.x ≤ y

D.x ≥ y

E.x=y or relationship between x and y cannot be determined

22) I: x² – 22x + 85 = 0

II: y² + 8y + 15 = 0

A.x < y

B.x > y

C.x ≤ y

D.x ≥ y

E.x=y or relationship between x and y cannot be determined

23) I: x²– 13x + 40 = 0

II: y² + y – 30 = 0

A.x < y

B.x > y

C.x ≤ y

D.x ≥ y

E.x=y or relationship between x and y cannot be determined

24) I: x²– 9x + 18 = 0

II: y² -9y + 20 = 0

A.x < y

B.x > y

C.x ≤ y

D.x ≥ y

E.x=y or relationship between x and y cannot be determined

25) I: x²+ 26x + 168 = 0

II: y² –52y + 667 = 0

A.x < y

B.x > y

C.x ≤ y

D.x ≥ y

E.x=y or relationship between x and y cannot be determined

26) I: x² – 18x – 63 = 0

II: y² + 7y – 120 = 0

A.x < y

B.x > y

C.x ≤ y

D.x ≥ y

E.x=y or relationship between x and y cannot be determined

27) I: 2x² + 3x – 20 = 0

II: 2y² – 13y + 20 = 0

A.x < y

B.x > y

C.x ≤ y

D.x ≥ y

E.x=y or relationship between x and y cannot be determined

28) I: 2x² – 19x + 24 = 0

II: y² – 17y + 60 = 0

A.x < y

B.x > y

C.x ≤ y

D.x ≥ y

E.x=y or relationship between x and y cannot be determined

29) I: x²+ 25x + 144 = 0

II: y² – 18y + 77 = 0

A.x < y

B.x > y

C.x ≤ y

D.x ≥ y

E.x=y or relationship between x and y cannot be determined

30) I: x²– 30x + 216 = 0

II: y² – y – 132 = 0

A.x < y

B.x > y

C.x ≤ y

D.x ≥ y

E.x=y or relationship between x and y cannot be determined

Answers :

21) Answer: A

x² + 6x + 8 = 0

=>x² + 2x +4x + 8 = 0

=> x(x + 2) +4(x + 2) = 0

=> (x + 2) (x + 4) = 0

=> x = -2, -4

y² – 8y + 15 = 0

=>y² – 3y -5y + 15 = 0

=> y(y – 3) -5(y – 3) = 0

=> (y – 3) (y – 5) = 0

=> y = 3, 5

Hence, x < y

22) Answer: B

x² – 22x + 85 = 0

=>x² – 17x – 5x + 85 = 0

=>x(x – 17) – 5(x – 17) = 0

=> (x – 5) (x – 17) = 0

=> x = 5, 17

y² + 8y + 15 = 0

=> y² + 5y + 3y + 15 = 0

=>y(y + 5) + 3(y + 5) = 0

=> (y + 3) (y + 5) = 0

=> y = -3, -5

Hence, x > y

23) Answer: D

x² – 13x + 40 = 0

=>x² – 5x -8x + 40 = 0

=>x(x – 5) -8(x – 5) = 0

=> (x – 5) (x – 8) = 0

=> x = 5, 8

y² + y – 30 = 0

=>y² -5y + 6y – 30 = 0

=>y(y – 5) +6(y – 5) = 0

=> (y – 5) (y + 6) = 0

=> y = 5, -6

Hence, x ≥ y

24) Answer: E

x² – 9x + 18 = 0

=>x² – 6x -3x + 18 = 0

=>x(x – 6) -3(x – 6) = 0

=> (x – 3) (x – 6) = 0

=> x = 3, 6

y² – 9y + 20 = 0

=>y² – 4y -5y+ 20 = 0

=>y(y – 4) -5(y – 4) = 0

=> (y – 4) (y – 5) = 0

=> y = 4, 5

Hence, relationship between x and y cannot be determined

25) Answer: A

x² + 26x + 168 = 0

=>x² + 12x + 14x + 168 = 0

=>x(x + 12) +14 (x + 12) = 0

=> (x + 12) (x + 14) = 0

=> x = – 12, – 14

y² – 52y + 667 = 0

=>y² – 29y – 23y + 667 = 0

=>y(y – 29) -23(y – 29) = 0

=> (y – 29) (y – 23) = 0

=> y = 29, 23

Hence, x < y

26) Answer: E

x² – 18x – 63 = 0

=>x² – 21x +3x – 63 = 0

=>x(x – 21) + 3(x – 21) = 0

=> (x + 3) (x – 21) = 0

=> x = -3, 21

y² + 7y – 120 = 0

=>y² + 15y – 8y – 120 = 0

=>y(y + 15) – 8(y + 15) = 0

=> (y – 8) (y + 15) = 0

=> y = 8, -15

Hence, relationship between x and y cannot be determined

27) Answer: C

2x² + 3x – 20 = 0

=> 2x(x + 4) – 5(x + 4) = 0

=> (2x – 5) (x + 4) = 0

=> x = 5/2, -4

2y² – 13y + 20 = 0

=> 2x(x – 4) – 5(x – 4) = 0

=> (2x – 5) (x – 4) = 0

=> x = 5/2, 4

Hence, x ≤ y

28) Answer: E

2x² – 19x + 24 = 0

=> (2x – 3) (x – 8) = 0

=> x = 3/2, 8

y² – 17y + 60 = 0

=> (y – 5) (y – 12) = 0

=> y = 5, 12

Hence, relationship between x and y cannot be determined

29) Answer: A

x² + 25x + 144 = 0

=> (x + 16) (x + 9) = 0

=> x = -16, -9

y² – 18y + 77 = 0

=> (y – 7) (y – 11) = 0

=> y = 7, 11

Hence, x < y

30) Answer: D

x² – 30x + 216 = 0

=> (x – 18) (x – 12) = 0

=> x = 12, 18

y² – y – 132 = 0

=> (y + 11) (y – 12) = 0

=> y = -11, 12

Hence, x ≥ y

 

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Directions (31-40): Following question contains two equations as I and II. You have to solve both equations and determine the relationship between them and give an answer as,

31) I) x² + x – 56 = 0

II)y²+ 20y + 96 = 0

A.x > y

B.x ≥ y

C.x = y or relationship can’t be determined.

D.x < y

E.x ≤ y

32) I) x² + 22x + 57 = 0

II)y²– 10y – 39 = 0

A.x > y

B.x ≥ y

C.x = y or relationship can’t be determined.

D.x < y

E.x ≤ y

33) I) 2x² – 2x – 12 = 0

II)y²+ 8y + 12 = 0

A.x > y

B.x ≥ y

C.x = y or relationship can’t be determined.

D.x < y

E.x ≤ y

34) I) 2x² – 18x + 36 = 0

II) y²–15y + 54 = 0

A.x > y

B.x ≥ y

C.x = y or relationship can’t be determined.

D.x < y

E.x ≤ y

35) I) x² + 23x + 90 = 0

II) y²– 8y – 48 = 0

A.x > y

B.x ≥ y

C.x = y or relationship can’t be determined.

D.x < y

E.x ≤ y

36) I) x² – 8x – 105 = 0

II)y²– 33y + 270 = 0

A.x > y

B.x ≥ y

C.x = y or relationship can’t be determined.

D.x < y

E.x ≤ y

37) I) x² – 10x + 16 =0

II)y²– 4y + 4 = 0

A.x > y

B.x ≥ y

C.x < y

D.x ≤ y

E.x = y or the relation cannot be established

38) I) x² + 10x + 25 =0

II)y²= 16

A.x > y

B.x ≥ y

C.x < y

D.x ≤ y

E.x = y or the relation cannot be established

39) I) x² – 18x + 32 = 0

II)y²+ 23y + 22 = 0

A.x > y

B.x ≥ y

C.x = y or relationship can’t be determined.

D.x < y

E.x ≤ y

40) I) x² + 14x + 40 = 0

II)y²+ 15y + 44 = 0

A.x > y

B.x ≥ y

C.x = y or relationship can’t be determined.

D.x < y

E.x ≤ y

Answers :

31) Answer: B

x² + x – 56 = 0

x² + 8x – 7x – 56 = 0

x(x + 8) – 7(x + 8) = 0

(x – 7)(x + 8) = 0

x = 7, -8

y² + 20y + 96 = 0

y² + 12y + 8y + 96 = 0

y(y + 12) + 8(y + 12) = 0

(y + 8)(y + 12) = 0

y = -8, -12

x ≥ y

32) Answer: E

x² + 22x + 57 = 0

x² + 19x + 3x + 57 = 0

x(x + 19) + 3(x + 19) = 0

(x + 3)(x + 19) = 0

x = -3, -19

y² – 10y – 39 = 0

y² – 13y + 3y – 39 = 0

y(y – 13) + 3(y – 13) = 0

(y + 3)(y – 13) = 0

y = -3, 13

x ≤ y

33) Answer: B

2x² – 2x – 12 = 0

2x² – 6x + 4x – 12 = 0

2x(x – 3)+ 4(x – 3) = 0

(2x + 4)(x – 3) = 0

x = -2, 3

y² + 8y + 12 = 0

y² + 6y + 2y + 12 = 0

y(y + 6) + 2(y + 6) = 0

(y + 2)(y + 6) = 0

y = -2, -6

x ≥ y

34) Answer: E

2x² – 18x + 36 = 0

2x² – 12x – 6x + 36 = 0

2x(x – 6) – 6(x – 6) = 0

(2x – 6)(x – 6) = 0

x = 3, 6

y² – 15y + 54 = 0

y² – 9y – 6y + 54 = 0

y(y – 9) – 6(y – 9) = 0

(y – 6)(y – 9) = 0

y =6, 9

Hence, x ≤ y

35) Answer: D

x² + 23x + 90 = 0

x² + 18x + 5x + 90 = 0

x(x + 18) + 5(x + 18) = 0

(x + 5)(x + 18) = 0

x = -5, -18

y² – 8y – 48 = 0

y² – 12y + 4y – 48 = 0

y(y – 12) + 4(y – 12) = 0

(y + 4)(y – 12) = 0

y = -4, 12

Hence, x < y

36) Answer: E

x² – 8x – 105 = 0

x² – 15x + 7x – 105 = 0

x(x – 15) + 7(x – 15) = 0

(x + 7)(x – 15) = 0

x =-7, 15

y² – 33y + 270 = 0

y² – 18y – 15y + 270 = 0

y(y – 18) – 15(y – 18) = 0

(y – 15)(y – 18) = 0

y = 15, 18

x ≤ y

37) Answer: B

x² – 10x + 16 =0

x² – 8x -2x +16 =0

x(x-8) -2( x-8) = 0

(x-8) (x-2) = 0

x =8, x= 2

y² – 4x + 4 = 0

y² -2y -2y +4 = 0

y(y-2) -2(y-2) = 0

(y-2) (y-2) = 0

y =2,2

x ≥ y

38) Answer: C

x² + 10x + 25 =0

x² + 5x +5x + 25 = 0

x(x+5) +5(x+5) = 0

(x+5) (x+5) = 0

x =-5,-5

y² = 16

y = +4, y = -4

x < y

39) Answer: A

x² – 18x + 32 = 0

x² – 16x – 2x + 32 = 0

x(x – 16) – 2(x – 16) = 0

(x – 2)(x – 16) = 0

x = 2, 16

y² + 23y + 22 = 0

y² + 22y + y + 22 = 0

y(y + 22) + 1(y + 22) = 0

(y + 1)(y + 22) = 0

y = -1, -22

x > y

40) Answer: C

x² + 14x + 40 = 0

x² + 10x + 4x + 40 = 0

x(x + 10) + 4(x + 10) = 0

(x + 4)(x + 10) = 0

x = -4, -10

y² + 15y + 44 = 0

y² + 11y + 4y + 44 = 0

y(y + 11) + 4(y + 11) = 0

(y + 4)(y + 11) = 0

y = -4, -11

Relationship between x and y cannot be established.

 

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