Quadratic Equation Question for IBPS Clerk 2017. Welcome to the letsstudytogether.co online Quant section. If you are preparing for IBPS PO/Clerk 2017 exam, you will come across a section on Quant Section.Quadratic Equation Quiz Set -4 IBPS Clerk.

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## Quadratic Equation Question for IBPS Clerk 2017

1.Directions: In the questions, two equations I and II are given. You have to solve both the equations and give answer

I. x2 + 5x + 6 = 0
II. (y2/2) – 4.5y + 10 = 0

(a) if x > y
(b) if x < y
(c) if x = y, or relation cannot be established between x and y
(d) if x ≥ y
(e) if x ≤ y

B. x < y

x2 + 5x + 6 =0
x2+ 2x + 3x +6 =0
(x + 2)(x + 3)
x = –2, –3
(y2/2) – 4.5y + 10 = 0
y2– 9y + 20 = 0

=> y2– 4y – 5y + 4*5 = 0
(y – 4)(y – 5)
y = 4, 5

x<y

2.Directions: In the questions, two equations I and II are given. You have to solve both the equations and give answer.

I. y2– 13y + 40 = 0
II. 6x2– 7x + 2 = 0

(a) if x > y
(b) if x < y
(c) if x = y, or relation cannot be established between x and y
(d) if x ≥ y
(e) if x ≤ y

B. x < y

y2– 13y + 40 = 0
y2– 8y – 5y + 5*8 = 0
y(y – 8) – 5(y – 8) = 0
(y – 5)(y – 8) = 0
y = 5, 8
6x2– 7x + 2 = 0
6x2– 4x – 3x + 2 = 0
2x(3x – 2) – (3x – 2) = 0
(2x – 1)(3x – 2) = 0
x = 1/2 or 2/3
x<y

3.Directions: In the questions, two equations I and II are given. You have to solve both the equations and give answer.

I. 15x2 + 57x – 12 = 0
II. 6y2+ 25y – 9 = 0

(a) if x > y
(b) if x < y
(c) if x = y, or relation cannot be established between x and y
(d) if x ≥ y
(e) if x ≤ y

C. if x = y, or relation cannot be established between x and y

15x2+ 57x – 12 = 0
15x2+ 60x – 3x – 12 = 0
15x(x + 4) – 3(x + 4) = 0
(15x – 3)(x + 4) = 0
3(5x – 1)(x + 4) = 0
x = 1/5, -4
6y2+ 25y – 9 = 0
6y2+ 27y – 2y – 9 = 0
3y(2y + 9) – (2y + 9) = 0
(3y – 1)(2y + 9)
y = 1/3 or -9/2
Therefore, relationship between x and y can’t be determined.

4.Directions: In the questions, two equations I and II are given. You have to solve both the equations and give answer.

I. 4x4– 5x2+ 1 = 0
II. y2– y – 2 = 0

(a) if x > y
(b) if x < y
(c) if x = y, or relation cannot be established between x and y
(d) if x ≥ y
(e) if x ≤ y

C. if x = y, or relation cannot be established between x and y

4x4– 5x2+ 1 = 0
4x4– 4x2– x2+ 1 = 0
4x2(x2– 1) – 1(x2– 1) = 0
(4x2– 1) (x2– 1) = 0
x = 1, -1, 1/2, -1/2
y2– y – 2 = 0
y2+ y – 2y – 2 = 0
y(y + 1) – 2 (y + 1) = 0
(y – 2) (y + 1) = 0
y = 2, -1
Therefore, relationship between x and y can’t be determined.

5.Directions: In the questions, two equations I and II are given. You have to solve both the equations and give answer.

I. 14y2+ 75y – 11 = 0
II. 6x2+ 7x + 2 = 0

(a) if x > y
(b) if x < y
(c) if x = y, or relation cannot be established between x and y
(d) if x ≥ y
(e) if x ≤ y

C. if x = y, or relation cannot be established between x and y

14y2+ 75y – 11 = 0
14y2+ 77y – 2y – 11 = 0
7y(2y + 11) – (2y + 11) = 0
(7y-1) (2y + 11)
y = 1/7, -11/2
6x2+ 7x + 2 = 0
6x2+ 3x + 4x + 2 = 0
3x(2x + 1) + 2(2x + 1) = 0
(3x + 2)(2x + 1) = 0
x = -2/3 or -1/2
Therefore, relationship between x and y can’t be determined.

6.Directions: In the questions, two equations I and II are given. You have to solve both the equations and give answer.

I. 2x2– 37x + 143 = 0
II. 3y2+ 31y – 22 = 0

(a) if x > y
(b) if x < y
(c) if x = y, or relation cannot be established between x and y
(d) if x ≥ y
(e) if x ≤ y

A. if x > y

2x2– 37x + 143 = 0
=> 2x2– 26x – 11x + 143 = 0
=> 2x(x-13) – 11(x-13) = 0
=> (x-13)(2x-11) = 0
=> x = 13, 11/2
3y2+ 31y – 22 = 0
3y2+ 33y – 2y – 22 = 0
=> 3y(y+11) – 2(y+11) = 0
=> (y+11)(3y-2)
=> y = -11, 2/3
Therefore, x > y

7.Directions: In the questions, two equations I and II are given. You have to solve both the equations and give answer.

I. 6x2 + 31x + 35 = 0
II. 4y2– 8y – 21 = 0

(a) if x > y
(b) if x < y
(c) if x = y, or relation cannot be established between x and y
(d) if x ≥ y
(e) if x ≤ y

B. if x < y

6×2 + 31x + 35 = 0
=> 6x2 + 21x + 10x + 35 = 0
=> 3x(2x+7) + 5(2x+7) = 0
=> (2x+7)(3x+5) = 0
x = -7/2, -5/3
4y2– 8y – 21 = 0
=> 4y2– 8y – 21 = 0
=> 4y2 + 6y – 14y – 21 = 0
=> (2y+3)(2y-7) = 0
=> y = -3/2, 7/2
Therefore, x < y

8.Directions: In the questions, two equations I and II are given. You have to solve both the equations and give answer.

I. 2x2– 11x – 21 = 0
II. y2– 3y – 28 = 0

(a) if x > y
(b) if x < y
(c) if x = y, or relation cannot be established between x and y
(d) if x ≥ y
(e) if x ≤ y

C. if x = y, or relation cannot be established between x and y

2x2– 11x – 21 = 0
=> 2x2– 14x + 3x – 21 = 0
=> 2x(x-7) + 3(x-7) = 0
=> (2x+3)(x-7) = 0
=> x = -3/2, 7
y2– 3y – 28 = 0
=> y2– 7y + 4y – 28 = 0
=> y(y-7) + 4(y-7) = 0
=> (y+4)(y-7) = 0
=> y = -4, 7
Therefore, no relation can be established between x and y.

9.Directions: In the questions, two equations I and II are given. You have to solve both the equations and give answer.

I. 35x2– 74x + 35 = 0
II. y2– y – 6 = 0

(a) if x > y
(b) if x < y
(c) if x = y, or relation cannot be established between x and y
(d) if x ≥ y
(e) if x ≤ y

C. if x = y, or relation cannot be established between x and y

35x2– 74x + 35 = 0
=> 35x2– 49x – 25x + 35 = 0
=> 7x(5x-7) – 5(5x-7) = 0
=> (7x-5)(5x-7) = 0
=> y = 5/7, 7/5
y2– y – 6 = 0
=> y2+ 2y – 3y – 6 = 0
=> y(y+2) – 3(y-2) = 0
=> (y+2)(y-3) = 0
=> y = – 2, 3
Therefore, relationship between x and y can’t be established

10.Directions: In the questions, two equations I and II are given. You have to solve both the equations and give answer.

I. x2+ 6x – 55 = 0
II. 3y2– 14y – 5 = 0

(a) if x > y
(b) if x < y
(c) if x = y, or relation cannot be established between x and y
(d) if x ≥ y
(e) if x ≤ y

C. if x = y, or relation cannot be established between x and y

x2+ 6x – 55 = 0
=> x2+ 11x – 5x – 55 = 0
=> x(x+11) – 5(x+11) = 0
=> (x+11)(x-5) = 0
=>x = -11, 5
3y2– 14y – 5 = 0
=>3y2+ y -15y – 5 = 0
=> y(3y+1) – 5(3y+1) = 0
=> (3y+1)(y-5) = 0
=> y = -1/3, 5
Therefore, no relationship between x and y can be established.