# Set-4 Quadratic Equations (Inequalities) For SBI PO and SBI Clerk 2019 | Must Go Through These Questions

Dear Aspirants,
We are providing the most important Quadratic Equations (Inequalities) for SBI PO 2019, SBI Clerk 2019 and all other competitive bank and insurance exams. These questions have very high chances to be asked in SBI PO 2019, SBI Clerk 2019.

Directions:1-10)In the following questions, two equations I and II are given. You have to solve both the equations and give Answer as,In the following questions, two equations I and II are given. You have to solve both the equations and give Answer as,

1.

1. √324 x + √11664 = 0
2. (16807)1/5 y + (9261)1/3 = 0

2.

1. 2x – y = 31
2. y = ∛19683

3.

1. x2 – 10x + 21 = 0
2. y2 – 10y + 16 = 0

4.

1. 12x2 – 17x – 57 = 0
2. 4y2 -7y – 36 = 0

5.

1. 3x + 5y = 34
2. 5x – 2y = 5

6.

1. x2 – 452 = 874 -√900
2. y = (24+√(121+√(500+29)))

7.

1. 3 + 18/x + 15/x2 =0

2. 2 + 4/y+2/y2 =0

8.

1. x3 = 512

2. y2 – 17y + 72 =0

9.

1. 9x+7y = 25

2. 5x+14y = 24

10.

1. 15x2 – 25x + 10 = 12x2-6x -6
2. 2y2 + 45y + 252 =0

• Directions:1-10)In the following questions, two equations I and II are given. You have to solve both the equations and give Answer as,In the following questions, two equations I and II are given. You have to solve both the equations and give Answer as,

1. √324 x + √11664 = 0
2. (16807)1/5 y + (9261)1/3 = 0

Ans:3

1. √324 x + √11664 = 0

18x + 108 = 0

18x = -108

x = – 6

2. (16807)1/5 y + (9261)1/3 = 0

(75)1/5y + (213)1/3 = 0

7y + 21 = 0

7y = -21

y = -3

Hence x < y

• ##### 2. Question
1. 2x – y = 31
2. y = ∛19683

Ans:1

1. y = ∛19683

y = 27

2. 2x – y = 31

2x – 27 = 31

2x = 58

x = 29

Hence x > y

• ##### 3. Question
1. x2 – 10x + 21 = 0
2. y2 – 10y + 16 = 0

Ans:5

1. x2 – 10x + 21 = 0

(x – 7) (x – 3) = 0

x = 7, 3

2. y2 – 10y + 16 = 0

(y -8) (y – 2) = 0

y = 8, 2

Hence the relationship can’t be determined.

• ##### 4. Question
1. 12x2 – 17x – 57 = 0
2. 4y2 -7y – 36 = 0

Ans:5

1. 12x2 – 17x – 57 = 0

12x2 – 36x + 19x – 57 = 0

12x(x – 3) + 19(x – 3) = 0

(12x + 19) (x – 3) = 0

x = -19/12, 3

x = -1.58, 3

2.4y2 -7y – 36 = 0

4y2 -16y + 9y -36 = 0

4y(y – 4) + 9(y – 4) = 0

(4y + 9) (y – 4) = 0

y = -9/4, 4

y= -2.25, 4

Hence the relationship can’t be determined.

• ##### 5. Question
1. 3x + 5y = 34
2. 5x – 2y = 5
Ans:3
3x + 5y = 34 –à (1)5x – 2y = 5—à (2)

By solving the equation (1) and (2), we get,

x = 3, y = 5

Hence x < y

• ##### 6. Question
1. x2 – 452 = 874 -√900
2. y = (24+√(121+√(500+29)))

Ans:4

1. x2 – 452 = 874 -√900

x2 = 1326 – 30 = 1296

x =± 36

2. y = (24+√(121+23))

y = (24+12)

y = 36

x ≤ y

y = (24+√(121+√(500+29)))

• ##### 7. Question
1. 3 + 18/x + 15/x2 =0

2. 2 + 4/y+2/y2 =0

Ans:4

1. 3 + 18/x + 15/x2 =0

3x2 + 18x + 15 =0

3x2 + 3x + 15x +15 =0

3x (x+1) +15 (x+1) =0

(3x+15) (x+1) =0

x =-1, -15/3 = -1, -5

2. 2y2 + 4y + 2 =0

2y2 +2y +2y +2 =0

2y (y+1) +2 (y+1) =0

(2y+2) (y+1) =0

y = -1, -2/2 = -1, -1

x ≤ y

2 + 4/y+2/y2 =0

• ##### 8. Question
1. x3 = 512

2. y2 – 17y + 72 =0

Ans:4

1. x3 = 512

x = 8

2. y2 – 8y – 9y+ 72 =0

(y-8) (y-9) =0

y = 8, 9

x ≤ y

y2 – 17y + 72 =0

• ##### 9. Question
1. 9x+7y = 25

2. 5x+14y = 24

Ans:1
9x+7y = 25 —– (1)5x+14y = 24 —– (2)

Simplify the above equation, we get x = 2 and y =1

x > y

• ##### 10. Question
1. 15x2 – 25x + 10 = 12x2-6x -6
2. 2y2 + 45y + 252 =0

Ans:1

1. 15x2 – 25x + 10 = 12x2-6x -6

3x2 – 19x + 16 =0

3x2 -3x – 16x + 16 =0

(3x-16) (x-1) =0

x=16/3, 1 = 5.33, 1

2. 2y2 + 24y +21y + 252 =0

2y (y+12) + 21 (y+12) =0

(2y+21) (y+12) =0

y= -21/2, -12 = -10.5, -12

x > y2y2 + 45y + 252 =0