# Aptitude Questions: Quadratic Equations Set 22

Hello Aspirants. Welcome to Online Quantitative Aptitude Section in AffairsCloud.com. Here we are creating sample questions in Quadratic Equations which is common for all the competitive exams. We have included Some questions that are repeatedly asked in bank exams !!!

1. x3 – 4x2 – 9x + 36 = 0
y3 + y2 + y – 3 = 0
A. X > Y
B. X < Y
C. X ≥ Y
D. X ≤ Y
E. X = Y or relation cannot be established
E. X = Y or relation cannot be established
Explanation:

x3 – 4x2 – 9x + 36 = 0
x2(x – 4) – 9(x – 4) = 0
(x2 – 9)(x – 4) = 0
(x – 3)(x + 3)(x – 4) = 0
x = 3, -3, 4
y3 + y2 + y – 3 = 0
(y – 1)(y2 + 2y + 3) = 0
y = 1; (y2 + 2y + 3) = 0 => No real Solution

2. x3 + 10x2 + 27x + 18 = 0
y3 + 9y2 + 23y + 15 = 0
A. X > Y
B. X < Y
C. X ≥ Y
D. X ≤ Y
E. X = Y or relation cannot be established
E. X = Y or relation cannot be established
Explanation:

x3 + 10x2 + 27x + 18 = 0
A3 + Bx2 + Cx + D = 0
D = Product, B = Sum
D = 18; Factors = 1, 2, 3, 6, 9, 18
Cubic equation has three roots, choose three factors(Sum of three factors = B) => 1, 3, 6
(x + 1)(x + 3)(x + 6) = 0
x = -1, -3, -6
y3 + 9y2 + 23y + 15 = 0
A3 + Bx2 + Cx + D = 0
D = Product, B = Sum
D = 15; Factors = 1, 3, 5, 15
Cubic equation has three roots, choose three factors(Sum of three factors = B) => 1, 3, 5
(y + 1)(y + 3)(y + 5) = 0
y = -1, -3, -5

3. x3 – 4x2 + 5x – 2 = 0
y³ + 7y2 + 14y + 8 = 0
A. X > Y
B. X < Y
C. X ≥ Y
D. X ≤ Y
E. X = Y or relation cannot be established
A. X > Y
Explanation:

x3 – 4x2 + 5x – 2 = 0
D = Product, B = Sum
D = 2; Factors = -1, -2
Cubic equation has three roots, choose three factors(Sum of three factors = B) => -1, -1, -2
(x – 1)(x – 1)(x – 2) = 0
x = 1, 1, 2
y³ + 7y2 + 14y + 8 = 0
D = 8; Factors = 1, 2, 4, 8
Cubic equation has three roots, choose three factors(Sum of three factors = B) => 1, 2, 4
(y + 1)(y + 2)(y + 4) = 0
y = -1, -2, -4

4. x2 – (16)2 = (23)2 – 56
y1/3 – 55 + 376 = (18)2
A. X > Y
B. X < Y
C. X ≥ Y
D. X ≤ Y
E. X = Y or relation cannot be established
D. X ≤ Y
Explanation:

x2 – (16)2 = (23)2 – 56
x2 = 729
x = ± 27
y1/3 – 55 + 376 = (18)2
y = 33 = 27

5. √36x + √64 = 0
√81y + (4)2 = 0
A. X > Y
B. X < Y
C. X ≥ Y
D. X ≤ Y
E. X = Y or relation cannot be established
A. X > Y
Explanation:

√36x + √64 = 0
6x + 8 = 0
x =  – 1.33
√81y + (4)2 = 0
9y + 16 = 0
y = – 1.77

6. 25/√x + 9/√x = 17√x
√y/3 + 5√y/6 = 3/√y
A. X > Y
B. X < Y
C. X ≥ Y
D. X ≤ Y
E. X = Y or relation cannot be established
B. X < Y
Explanation:

25/√x + 9/√x = 17√x
34 = 17x
x = 2
21√y/18 = 3/√y
y = 18/7 = 2.57

7. (625)1/4x + √1225 = 155
√196y + 13 = 279
A. X > Y
B. X < Y
C. X ≥ Y
D. X ≤ Y
E. X = Y or relation cannot be established
A. X > Y
Explanation:

(625)1/4x + √1225 = 155
5x = 155 – 35
5x = 120 => x = 24
√196y + 13 = 279
14y = 266
y = 19

8. 3x – 2y = 10
5x – 6y = 6
A. X > Y
B. X < Y
C. X ≥ Y
D. X ≤ Y
E. X = Y or relation cannot be established
A. X > Y
Explanation:

3x – 2y = 10 — (1)
5x – 6y = 6 —(2)
From eqn (1) and (2)
x = 6; y = 4

9. √x – √6/√x = 0
y3 – 6(3/2) = 0
A. X > Y
B. X < Y
C. X ≥ Y
D. X ≤ Y
E. X = Y or relation cannot be established
E. X = Y or relation cannot be established
Explanation:

√x – √6/√x = 0
x = √6 = 2.44
y3 – 6(3/2) = 0
y = 6(1/2)
y = √6 = 2.44

10. √(x + 6) = √121 – √36
y2 + 112 = 473
A. X > Y
B. X < Y
C. X ≥ Y
D. X ≤ Y
E. X = Y or relation cannot be established
C. X ≥ Y
Explanation:

√(x + 6) = √121 – √36
√(x + 6) = 11 – 6
√(x + 6) = 5
x + 6 = 25
x = 19
y2 + 112 = 473
y2 = 361
y = ± 19

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