Quadratic Equations Practice Sets SBI PO, IBPS PO, NIACL, NICL, LIC, Dena Bank PO PGDBF, BOI, Bank of Baroda and other competitive exams
Directions: In the following questions, two equations numbered are given in variables x and y. You have to solve both the equations and find out the relationship between x and y. Then give answer accordingly-
- I. 3x2 – 13x – 10 = 0,
II. 2y2 + 11y + 12 = 0
A) If x > y
B) If x < y
C) If x ≥ y
D) If x ≤ y
E) If x = y or relation cannot be established
View Answer
Option A
Solution:
3x2 – 13x – 10 = 0
3x2 – 15x + 2x – 10 = 0
Gives x = -2/3, 5
2y2 + 11y + 12 = 0
2y2 + 8y + 3y + 12 = 0
Gives y = -4, -3/2
Put all values on number line and analyze the relationship
-4…. -3/2….-2/3….. 5
- I. 3x2 – 8x – 16 = 0,
II. 3y2 – 26y + 56 = 0
A) If x > y
B) If x < y
C) If x ≥ y
D) If x ≤ y
E) If x = y or relation cannot be established
View Answer
Option D
Solution:
3x2 – 8x – 16 = 0
3x2 – 12x + 4x – 16 = 0
Gives x = -4/3, 4
3y2 – 26y + 56 = 0
3y2 – 26y + 56 = 0
Gives y = 4, 14/3
- I. 3x2 + 10x – 8 = 0,
II. 3y2 + 10y + 8 = 0
A) If x > y
B) If x < y
C) If x ≥ y
D) If x ≤ y
E) If x = y or relation cannot be established
View Answer
Option E
Solution:
3x2 + 10x – 8 = 0
3x2 + 12x – 2x – 8 = 0
Gives x = -2, 2/3
3y2 + 10y + 8 = 0
3y2 + 6y + 4y + 8 = 0
Gives y = -2, -4/3
- I. 3x2 – 25x + 52 = 0,
II. 2y2 – 17y + 36 = 0
A) If x > y
B) If x < y
C) If x ≥ y
D) If x ≤ y
E) If x = y or relation cannot be established
View Answer
Option E
Solution:
3x2 – 25x + 52 = 0
3x2 – 12x – 13x + 52 = 0
Gives x = 4, 13/3
2y2 – 17y + 36 = 0
2y2 – 8y – 9y + 36 = 0
Gives y= 4, 9/2
- I. 3x2 + 7x – 6 = 0,
II. 4y2 – 11y + 6 = 0
A) If x > y
B) If x < y
C) If x ≥ y
D) If x ≤ y
E) If x = y or relation cannot be established
View Answer
Option B
Solution:
3x2 + 7x – 6 = 0
3x2 + 9x – 2x – 6 = 0
Gives x = -3, 2/3
4y2 – 11y + 6 = 0
4y2 – 8y – 3y + 6 = 0
Gives y= 3/4, 2
- I. 2x2 – 3x – 14 = 0,
II. 3y2 + 16y + 20 = 0
A) If x > y
B) If x < y
C) If x ≥ y
D) If x ≤ y
E) If x = y or relation cannot be established
View Answer
Option C
Solution:
2x2 – 3x – 14 = 0
2x2 + 4x – 7x – 14 = 0
Gives x = -2, 7/2
3y2 + 16y + 20 = 0
3y2 + 6y + 10y + 20 = 0
Gives y = -10/3, -2
- I. 7x2 + 19x – 6 = 0,
II. 3y2 – 8y – 16 = 0
A) x > y
B) x < y
C) x ≥ y
D) x ≤ y
E) x = y or relationship cannot be determined
View Answer
Option E
Solution:
7x2 + 19x – 6 = 0
7x2 + 21x – 2x – 6 = 0
Gives x = -3, 2/7
3y2 – 8y – 16 = 0
3y2 – 12y + 4y – 16 = 0
So y = -4/3, 4
- I. 8x2 + (4 + 2√2)x + √2 = 0
II. 3y2 – (6 + 2√3)y + 4√3 = 0
A) If x > y
B) If x < y
C) If x ≥ y
D) If x ≤ y
E) If x = y or relation cannot be established
View Answer
Option B
Solution:
8x2 + (4 + 2√2)x + √2 = 0
(8x2 + 4x) + (2√2x + √2) = 0
4x (2x + 1) + √2 (2x + 1) = 0
So x = -1/2 (-0.5), -√2/4 (-0.35)
3y2 – (6 + 2√3)y + 4√3 = 0
(3y2 – 6y) – (2√3y – 4√3) = 0
3y (y – 2) – 2√3 (y – 2) = 0
So, y = 2, 2√3/3
- I. 3x2 – (3 – 2√2)x – 2√2 = 0
II. 5y2 + (2 + 5√2)y + 2√2 = 0
A) If x > y
B) If x < y
C) If x ≥ y
D) If x ≤ y
E) If x = y or relation cannot be established
View Answer
Option E
Solution:
3x2 – (3 – 2√2)x – 3√2 = 0
(3x2 – 3x) + (2√2x – 2√2) = 0
3x (x – 1) + 2√2 (x – 1) = 0
So x = 1, -2√2/3 (-0.9)
5y2 + (2 + 5√2)y + 2√2 = 0
(5y2 + 2y) + (5√2y + 2√2) = 0
y (5y + 2) + √2 (5y + 2) = 0
So, y = -2/5 (-0.4), -√2 (-1.4)
- I. 6x2 – (3 + 4√3)x + 2√3 = 0,
II. y2 – (3 + √3)y + 3√3 = 0
A) x > y
B) x < y
C) x ≥ y
D) x ≤ y
E) x = y or relationship cannot be determined
View Answer
Option B
Solution:
6x2 – (3 + 4√3)x + 2√3 = 0
(6x2 – 3x) – (4√3x – 2√3) = 0
3x (2x- 1) – 2√3 (x – 2) = 0,
So x = 1/2, 2√3/3 (1.15)
y2 – (3 + √3)y + 3√3 = 0
(y2 – 3y) – (√3y – 3√3) = 0
y (y – 3) – √3 (y – 3) = 0
So x = 3, √3 (1.73)
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