The official SBI PO Notification 2024 is expected to be released soon, and for those preparing for the exam, it is essential to focus on the key areas. One such important topic is Quadratic Equations, which comes in the examination every year. It is a scoring subject and you can solve all the questions with little effort. To strengthen your preparation, we have prepared a series of quadratic equation questions for SBI PO exam. Solutions to these questions are given in detail, so that you can improve your understanding and prepare for the exam with confidence.
Quadratic Equation Questions for SBI PO Exam
Quadratic Equations is an important and scoring topic in SBI PO 2024 exam from which 5 to 6 questions are asked every year. The Quantitative Aptitude section consists of a total of 35 questions covering various topics like Arithmetic, Number Series, Estimation, Data Interpretation (DI) and Quadratic Equations. If you pay special attention to quadratic equations and practice regularly, it will help you in increasing your marks and strengthen your preparation to get good results in the exam.
Directions (1-5): In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give
Answer
(a) if x>y
(b) if x≥y
(c) if x
(d) if x≠y
(e) If x = y or no relation can be established between x and y
Q1. I. x2 + 13x – 114 = 0
II. y3 = 216
Q2. ix2 − 6x + 12 = 4
II.2 + 4y − 10 = −13
Q3. I. 12x2 − 7x + 1 = 0
II. 20 years2 – 9y + 1 = 0
Q4. I. x2 + 26x + 165 = 0
II.2 + 23y + 132 = 0
Q5. I. x2 + x − 6 = 0
II. 15 years2 − 11y + 2 = 0
Direction (6 – 10): In each of the following questions two equations are given. Solve these equations and answer:
(a) If 𝑥 ≥ 𝑦, i.e., 𝑥 is greater than or equal to 𝑦
(b) If 𝑥 > 𝑦, i.e., 𝑥 is greater than 𝑦
(c) if 𝑥 ≤ 𝑦, i.e., 𝑥 is less than or equal to 𝑦
(d) if 𝑥 < 𝑦, i.e., 𝑥 is less than 𝑦
(e) 𝑥 = 𝑦 or no relation can be established between 𝑥 and 𝑦
Q6. (i) x2 + 9 = 73
(ii) y3 = 512
Q7. (i) x2 + 11x + 18 = 0
(ii) Y2 + 19y + 90 = 0
Q8. (And). 𝑥2 − 10𝑥 + 21 = 0
(ii). 𝑦² – 5𝑦 + 6 = 0
Q9. (i) 2x2 + x − 1 = 0
(ii) 2 years2 + 3y + 1 =0
Q10. (I). 2x² + 13x + 21 = 0
(ii). 2y² + 11y + 14 = 0
Direction (11 – 15): In each of these questions, two equations (I) and (II) are given.
You have to solve both the equations and give the answer.
(a) If x=y or no relation can be established.
(b) if x>y
(c) if x
(d) if x≥y
(e) if x≠y
Question 11. (I) x3 – 12 – 1319 = 0
(ii) y2 – 21 – 100 = 0
Question 12. i.12𝑥2 − 7𝑥 + 1 = 0
II. b2 + 23𝑦 + 132 = 0
Q13. (i) x2+ 9x – 52 = 0
(II) 12 years2 + 16y + 4 = 0
Question 14. (i) x2 – x – 210 = 0
(ii) y2 – 31y + 240 = 0
Question 15. (i) 2x2 – 8x – 24 = 0
(II) 9 years2 – 12y + 4 = 0
Directions (16-20): In each question two equations (I) and (II) are given. You must solve both equations and mark
Appropriate answer.
(a) if x > y
(b) if x ≥ y
(c) if x < y
(d) if x ≤ y
(e) If = y or the relation cannot be established.
Q16. I. 2x² – 7x + 5 = 0
II. y2 – 3y + 2 = 0
Q17. I. x² – 25x + 156 = 0
II. y² – 29y + 210 = 0
Question 18. I. x² + 20x + 96 = 0
II. y2 + 15y + 56 = 0
Question 19. I. x² – 3x – 40 = 0
II. 2y² + 11y + 15 = 0
Q20. I. x² – 16x + 64 = 0
II. y2 – 14y + 48 = 0
Solution
S1. Answer(D)
Sol. I. x² + 13x – 114 = 0
x²+ 19x – 6x – 114 =0
x(x + 19) – 6(x + 19) = 0
(x + 19) (x – 6) = 0
x = – 19, 6
II.3 = 216
this3 = 63
y = 6
So, 𝑥 ≤ 𝑦, that is, 𝑥 is less than or equal to 𝑦
S2. Answer(A)
Sol. I. 𝑥
2 − 6𝑥 + 8 = 0
𝑥2 − 2𝑥 − 4𝑥 + 8 = 0
𝑥(𝑥 − 2) − 4(𝑥 − 2) = 0
(𝑥 − 2)(𝑥 − 4) = 0
𝑥 = 2, 4
II. b2 + 4𝑦 + 3 = 0
𝑦2 + 𝑦 + 3𝑦 + 3 = 0
𝑦(𝑦 + 1) + 3(𝑦 + 1) = 0
(𝑦 + 1)(𝑦 + 3) = 0
𝑦 = −1, −3
So, 𝑥 > 𝑦
S3. Answer (B) Sol. i.12𝑥
2 − 7𝑥 + 1 = 0
12𝑥2 − 4𝑥 − 3𝑥 + 1 = 0
4𝑥(3𝑥 − 1) − 1(3𝑥 − 1) = 0
(3𝑥 − 1) (4𝑥 − 1) = 0
𝑥=1/3, 1/4
II. 20𝑦2 − 9𝑦 + 1 = 0
20𝑦2 − 5𝑦 − 4𝑦 + 1 = 0
5𝑦(4𝑦 − 1) − 1(4𝑦 − 1) = 0
(4𝑦 – 1) (5𝑦 – 1) = 0
𝑦 = 1/4, 1/5
So, 𝑥 ≥ 𝑦
S4. Ans.(e)
Sol. I. 𝑥2 + 26𝑥 + 165 = 0
𝑥2 + 11𝑥 + 15𝑥 + 165 = 0
𝑥(𝑥 + 11) + 15(𝑥 + 11) = 0
(𝑥 + 11)(𝑥 + 15) = 0
𝑥 = −11, −15
II. b2 + 23𝑦 + 132 = 0
𝑦2 + 11𝑦 + 12𝑦 + 132 = 0
𝑦(𝑦 + 11) + 12(𝑦 + 11) = 0
(𝑦 + 11)(𝑦 + 12) = 0
𝑦 = −11, −12
Hence no relation can be established
S6. Answer (C)
Sol. (i) x2 = 73−9
x = ± 8
(ii) y = + 8
So, 𝑥 ≤ 𝑦
S7. Answer(A)
Sol. (i) x2 + 9x + 2x + 18 = 0
x(x + 9) + 2(x + 9) = 0
x = −2, −9
(ii) Y2 + 10y + 9y + 90 =0
y(y + 10) + 9(y + 10) =0
y = −9, −10
So, 𝑥 ≥ 𝑦
S8. Answer(A)
Sol. (I). 𝑥2 − 3𝑥 − 7𝑥 + 21 = 0
𝑥(𝑥 − 3) − 7(𝑥 − 3) = 0
(𝑥 − 7)(𝑥 − 3) = 0
𝑥 = 7, 3
(ii). b2– 3𝑦 – 2𝑦 + 6 = 0
𝑦(𝑦 − 3) − 2(𝑦 − 3) = 0
(𝑦 − 3)(𝑦 − 2) = 0
𝑦 = 3, 2
So, 𝑥 ≥ 𝑦
S9. Answer(E)
Sol. (i) 2x2 + x − 1 = 0
2×2 + 2x – x – 1 = 0
2x(x + 1) −1(𝑥 + 1) = 0
x=1/2, −1
(ii) 2y2 + 3y + 1 =0
2y2 + 2y + y + 1 =0
2y(y + 1)+ 1(y + 1) = 0
y = −1, −1/2
Therefore, no relation can be established between 𝑥 and y
S10. Answer(E)
Sol. (I). 2×2 + 7x+6x+21 = 0
x(2x+7)+3(2x+7) = 0
(2x+7)(x+3) = 0
x = -3, -3.5
(ii). 2y² + 7y+4y + 14 = 0
y(2y+7) +2(2y+7) = 0
y = -2, -3.5
Hence, no relation can be established between 𝑥 and 𝑦
S11. Answer(D)
Sol. I. x=11
II. y=±11
So, x≥y
S12. Answer(B)
Sol. I. 12𝑥2 − 7𝑥 + 1 = 0
12𝑥2 − 4𝑥 − 3𝑥 + 1 = 0
4𝑥(3𝑥 − 1) − 1(3𝑥 − 1) = 0
(3𝑥 − 1) (4𝑥 − 1) = 0
𝑥 = 1/3, 1/4
II. b2 + 23𝑦 + 132 = 0
𝑦2 + 11𝑦 + 12𝑦 + 132 = 0
𝑦(𝑦 + 11) + 12(𝑦 + 11) = 0
(𝑦 + 11)(𝑦 + 12) = 0
𝑦 = −11, −12
So, x>y
S13. Answer(A)
Sol. I. x2+ 9x – 52 = 0
x2+ 13x – 4x – 52 = 0
x(x+13)-4(x+13) =0
(x+13) (x – 4) =0
x= −13, 4
II. 12y2 + 16y + 4 = 0
12y2 + 12y+4y + 4 = 0
12y(y+1)+4(y+1)=0
(12y+4)(y+1)=0
y=-1/3,-1
Hence no relation can be established.
S14. Year
Sol. I. x2 – x – 210 = 0
x2 –15x+14x – 210 = 0
x(x – 15)+14(x – 15)=0
(x+14) (x – 15) =0
x=-14, 15
II.2 – 31y + 240 = 0
this2 – 16y -15y+ 240 = 0
y(y – 16) – 15(y – 16)=0
(Y – 16) (Y – 15) =0
y=16, 15
So, x≠y
S15. Answer(A)
Sol. I. 2×2 – 8x – 24 = 0
𝑥2 − 4𝑥 − 12 = 0
𝑥2 − 6𝑥 + 2𝑥 − 12 = 0
𝑥 = 6, −2
II. year 92 – 12y + 4 = 0
year 92 −6y−6y + 4 = 0
3y(3y– 2) – 2(3y– 2)=0
(3y-2) (3y-2) =0
y=2/3, 2/3
Hence no relation can be established.
S16. Answer(E)
Sol. I. 2x² – 7x + 5 = 0
2x² – 2x – 5x + 5 = 0
(x – 1) (2x – 5) = 0
x = 1, 2.5
II. y2 – 3y + 2 = 0
y² − y – 2y + 2 = 0
(y – 1) (y – 2) = 0
y = 1, 2
∴ 𝑛𝑜 𝑟𝑒𝑙𝑎𝑡𝑖𝑜𝑛
S17. Answer (C)
Sol. I. x² – 25x + 156 = 0
x² – 12x – 13x + 156 = 0
(x – 12) (x – 13) = 0
x = 12, 13
II. y² – 29y + 210 = 0
y² – 14y – 15y + 210 = 0
(y – 14) (y – 15) = 0
y = 14, 15
∴ 𝑦 > 𝑥
S18. Answer(D)
Sol. I. x² + 20x + 96 = 0
x² + 8x + 12x + 96 = 0
(x + 8) (x + 12) = 0
x = -8, -12
II. y2 + 15y + 56 = 0
y² + 7y + 8y + 56 = 0
(y + 7) (y + 8) = 0
y = -7, -8
∴ x ≤ y
S19. Answer(E)
Sol. I. x² – 3x – 40 = 0
x² – 8x + 5x – 40 = 0
x(x – 8) + 5(x – 8) = 0
(x – 8) (x + 5) = 0
x = 8, -5
II. 2y² + 11y + 15 = 0
2y² + 6y + 5y + 15 = 0
2y (y + 3) + 5 (y + 3) =0
(2y + 5) (y + 3) =0
y = –5/2, –3
Therefore the connection is not able to be established.
