# Quadratic Equations Set – 16

1. I. 4x2 + 5x – 6 = 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

Option E
Solution:

4x2 + 5x – 6 = 0
4x2 + 8x – 3x – 6 = 0
Gives x = -2, 3/4
2y2 + 11y + 12 = 0
2y2 + 8y + 3y + 12 = 0
Gives y = -4, -3/2

2. I. 12x2 – 49x + 30 = 0,
II. 6y2 – 35y + 50 = 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

Option E
Solution:

12x2 – 49x + 30 = 0
12x2 – 9x – 40x + 30 = 0
Gives x = 3/4, 10/3
6y2 – 35y + 50 = 0
6y2 – 15y – 20y + 50 = 0
Gives y = 5/2, 10/3

3. I. 4x2 + 13x + 10 = 0,
II. 4y2 – 7y – 15 = 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

Option D
Solution:

4x2 + 13x + 10 = 0
4x2 + 8x + 5x + 10 = 0
Gives x = -2, -5/4
4y2 – 7y – 15 = 0
4y2 – 12y + 5y – 15 = 0
Gives y = -5/4, 3

4. I. 12x2 – 5x – 3 = 0,
II. 6y2 + y – 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

Option E
Solution:

12x2 – 5x – 3 = 0
12x2 + 4x – 9x – 3 = 0
Gives x = -1/3, 3/4
6y2 + y – 2 = 0
6y2 – 3y + 4y – 2 = 0
Gives y= -2/3, 1/2

5. I. 3x2 + 7x – 6 = 0,
II. 3y2 – 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

Option D
Solution:

Explanation:
3x2 + 7x – 6 = 0
3x2 + 9x – 2x – 6 = 0
Gives x = -3, 2/3
3y2 – 11y + 6 = 0
3y2 – 9y – 2y + 6 = 0
Gives y = 2/3, 3

6. I. 5x2 – 36x – 32 = 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

Option A
Solution:

5x2 – 36x – 32 = 0
5x2 + 4x – 40x – 32 = 0
Gives x = -4/5, 8
3y2 + 16y + 20 = 0
3y2 + 6y + 10y + 20 = 0
Gives y = -10/3, -2

7. I. 3x2 – (6 + 2√3)x + 4√3 = 0,
II. 3y2 – (2 + 3√3)y + 2√3 = 0
A) x > y
B) x < y
C) x ≥ y
D) x ≤ y
E) x = y or relationship cannot be determined

Option E
Solution:

3x2 – (6 + 2√3)x + 4√3 = 0
(3x2 – 6x) – (2√3x – 4√3) = 0
3x (x- 2) – 2√3 (x – 2) = 0,
So x = 2, 2√3/3 (1.15)
3y2 – (2 + 3√3)y + 2√3 = 0
(3y2 – 2y) – (3√3y – 2√3) = 0
y (3y – 2) – √3 (3y – 2) = 0
So x = 2/3, √3 (1.73)

8. I. 2x2 + (4 + √2)x + 2√2 = 0
II. y2 – (1 + 3√3)y + 3√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

Option B
Solution:

2x2 + (4 + √2)x + 2√2 = 0
(2x2 + 4x) + (√2x + 2√2) = 0
2x (x + 2) + √2 (x + 2) = 0
So x = -2, -√2/2 (-0.7)
y2 – (1 + 3√3)y + 3√3 = 0
(y2 – y) – (3√3y – 3√3) = 0
y (y – 1) – 3√3 (y – 1) = 0
So, y = 1, 3√3 (5.2)

9. I. x2 + (3 + 2√2)x + 6√2 = 0
II. 5y2 – (1 + 5√2)y + √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

Option B
Solution:

x2 + (3 + 2√2)x + 6√2 = 0
(x2 + 3x) + (2√2x + 6√2) = 0
x (x + 3) + 2√2 (x + 3) = 0
So x = -3, -2√2
5y2 – (1 + 5√2)y + √2 = 0
(5y2 – y) – (5√2y – √2) = 0
y (5y – 1) – 3√2 (5y – 1) = 0
So, y = 1/5, 3√2

10. I. 2x2 + (4 + 2√6)x + 4√6 = 0
II. 5y2 + (10 + √6)y + 2√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