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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.
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Directions:1-5) In the following questions, two equations numbered are given in variables a and b. You have to solve both the equations and find out the relationship between a and b. Then give answer accordingly-
1.
I.3a2 – 8a – 28 = 0
II.3b2 – 44b + 140 = 0
2.
I.2a2 – 27a + 81 = 0
II.3b2 – 50b + 168 = 0
3.
I.2a2 – 27a + 81 = 0
II.3b2 – 50b + 168 = 0
4.
I.3a2 – 14a – 24 = 0
II.3b2 + 32b + 64 = 0
5.
I.4a2 + a – 18 = 0
II.4b2 – 11b – 45 = 0
Directions:6-10) In each of these questions, two equations numbered I and II with variables x and y are given. You have to solve both the equations to find the relation between x and y.
6.
I. 5x2 – 176x – 333 = 0
II. 49y2 – 84y + 36 = 0
7.
I. x2 + 13√2x + 84 = 0
II. 12y2 – 20y + 8 = 0
8.
I. 20x2 – 108x + 144 = 0
II. y2 – 26y + 168 = 0
9.
I. x2 – 32x + 252 = 0
II. 25y2 – 90y + 72 = 0
10.
I. x2 – 56x + 783 = 0
II. 12y2 + 82y + 140 = 0
Check your Answers below:
Directions:1-5) In the following questions, two equations numbered are given in variables a and b. You have to solve both the equations and find out the relationship between a and b. Then give answer accordingly-
1. Question
I.3a2 – 8a – 28 = 0
II.3b2 – 44b + 140 = 0
Ans:4
3a2 – 8a – 28 = 0
3a2 + 6a – 14a – 28 = 0
Gives a = -2, 14/3
3b2 – 44b + 140 = 0
3b2 – 30b -14b + 140 = 0
Gives b = 14/3, 10
a ≤ b2. Question
I.2a2 – 27a + 81 = 0
II.3b2 – 50b + 168 = 0Ans:4
2a2 – 27a + 81 = 0
2a2 – 18a – 9a + 81 = 0
Gives a = 9/2, 9
3b2 – 50b + 168 = 0
3b2 – 36b – 14b + 168 = 0
Gives b = 14/3, 12
Cannot be determined3. Question
I.2a2 – 27a + 81 = 0
II.3b2 – 50b + 168 = 0
Ans:5
2a2 – 27a + 81 = 0
2a2 – 18a – 9a + 81 = 0
Gives a = 9/2, 9
3b2 – 50b + 168 = 0
3b2 – 36b – 14b + 168 = 0
Gives b = 14/3, 12
Cannot be determined4. Question
I.3a2 – 14a – 24 = 0
II.3b2 + 32b + 64 = 0Ans:1
3a2 – 14a – 24 = 0
3a2 – 18a + 4a – 24 = 0
Gives a = -4/3, 6
3b2 + 32b + 64 = 0
3b2 + 24b + 8b + 64 = 0
Gives b= -8, -8/3
a > b5. Question
I.4a2 + a – 18 = 0
II.4b2 – 11b – 45 = 0Ans:5
4a2 + a – 18 = 0
4a2 – 8a + 9a – 18 = 0
Gives a = -9/4, 2
4b2 – 11b – 45 = 0
4b2 – 20b + 9b – 45 = 0
Gives b= -9/4, 5Cannot be determined
Directions:6-10) In each of these questions, two equations numbered I and II with variables x and y are given. You have to solve both the equations to find the relation between x and y.
6. Question
I. 5x2 – 176x – 333 = 0
II. 49y2 – 84y + 36 = 0
Ans:5
I. 5x2 – 176x – 333 = 0=> (x – 37)(5x + 9) = 0
=> x = 37, -9/5
II. 49y2 – 84y + 36 = 0
=> (7y – 6)2 = 0
=>7y – 6 = 0
=> y = 6/7
Hence, relationship between x and y cannot be determined.
7. Question
I. x2 + 13√2x + 84 = 0
II. 12y2 – 20y + 8 = 0
Ans:3
I. x2 + 13√2x + 84 = 0=> (x + 6√2)(x + 7√2) = 0
=> x = -6√2, -7√2
II. 12y2 – 20y + 8 = 0
=> 3y2 – 5y + 2 = 0
=> (3y – 2)(y – 1) = 0
=> y = 2/3, 1
Hence, x < y
8. Question
I. 20x2 – 108x + 144 = 0
II. y2 – 26y + 168 = 0
Ans:3
I. 20x2 – 108x + 144 = 0=>5x2 – 27x + 36 = 0
=> (5x – 12)(x – 3) = 0
=> x = 12/5, 3
II. y2 – 26y + 168 = 0
=> (y – 12)(y – 14) = 0
=> y = 12, 14
Hence, x < y
9. Question
I. x2 – 32x + 252 = 0
II. 25y2 – 90y + 72 = 0
Ans:1
I. x2 – 32x + 252 = 0=> (x – 18)(x – 14) = 0
=> x = 18, 14
II. 25y2 – 90y + 72 = 0
=> (5y – 12)(5y – 6) = 0
=> y = 12/5, 6/5
Hence, x > y
10. Question
I. x2 – 56x + 783 = 0
II. 12y2 + 82y + 140 = 0
Ans:1
I. x2 – 56x + 783 = 0=> (x – 29)(x – 27) = 0
=> x = 29, 27
II. 12y2 + 82y + 140 = 0
=> 6y2 + 41y + 70 = 0
=> (2y + 7)(3y + 10) = 0
=> y = -7/2, -10/3
Hence, x > y
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