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 !!!
Follow the link To solve Quadratic Equations with the help of Number Line
- 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 establishedE. 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 - 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 establishedE. 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 - 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 establishedA. 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 - 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 establishedD. X ≤ Y
Explanation:
x2 – (16)2 = (23)2 – 56
x2 = 729
x = ± 27
y1/3 – 55 + 376 = (18)2
y = 33 = 27 - √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 establishedA. X > Y
Explanation:
√36x + √64 = 0
6x + 8 = 0
x = – 1.33
√81y + (4)2 = 0
9y + 16 = 0
y = – 1.77 - 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 establishedB. X < Y
Explanation:
25/√x + 9/√x = 17√x
34 = 17x
x = 2
21√y/18 = 3/√y
y = 18/7 = 2.57 - (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 establishedA. X > Y
Explanation:
(625)1/4x + √1225 = 155
5x = 155 – 35
5x = 120 => x = 24
√196y + 13 = 279
14y = 266
y = 19 - 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 establishedA. X > Y
Explanation:
3x – 2y = 10 — (1)
5x – 6y = 6 —(2)
From eqn (1) and (2)
x = 6; y = 4 - √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 establishedE. 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 - √(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 establishedC. 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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