1. I. 3x2 + 10x – 8 = 0,
II. 3y2 – 20y + 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
Answer & Explanation Option D
Solution:
3x2 + 10x – 8 = 0
3x2 + 12x – 2x + 8 = 0
Gives x = -4, 2/3
3y2 – 20y + 12 = 0
3y2 – 18y – 2y + 12 = 0
Gives y = 2/3, 6
2. I. 4x2 – 23x + 15 = 0,
II. 4y2 + 9y – 9 = 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
Answer & Explanation Option C
Solution:
4x2 – 23x + 15 = 0
4x2 – 20x -3x + 15 = 0
Gives x = 3/4, 5
4y2 + 9y – 9 = 0
4y2 + 12y – 3y – 9 = 0
Gives y = -3, 3/4
3. I. 5x2 – 13x – 6 = 0,
II. 5y2 – 18y – 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
Answer & Explanation Option E
Solution:
x = -2/5, 3
y = -2/5, 4
4. I. 2x2 + 7x – 4 = 0,
II. 3y2 – 19y + 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
Answer & Explanation Option B
Solution:
2x2 + 7x – 4 = 0
Gives x = -4, 1/2
3y2 – 19y + 20 = 0
Gives y= 4/3, 5
5. I. 2x2 – 3x – 9 = 0,
II. 3y2 + 13y + 14 = 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
Answer & Explanation Option A
Solution:
2x2 – 3x – 9 = 0
Gives x = -3/2, 3
3y2 + 13y + 14 = 0
Gives y= -7/3, -2
6. I. 3x2 – x – 10 = 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
Answer & Explanation Option E
Solution:
3x2 – x – 10 = 0
Gives x = -5/3, 2
3y2 – 11y + 6 = 0
Gives y = 2/3, 3
7. I. 3x2 – (3 – 2√2)x – 2√2 = 0
II. 3y2 – (1 + 3√3)y + √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
Answer & Explanation Option E
Solution:
3x2 – (3 – 2√2)x – 2√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)
3y2 – (1 + 3√3)y + √3 = 0
(3y2 – y) – (3√3y – √3) = 0
y (3y – 1) – √3 (3y – 1) = 0
So, y = 1/3, √3 (1.7)
8. I. x2 + (4 + √2)x + 4√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
Answer & Explanation Option D
Solution:
x2 + (4 + √2)x + 4√2 = 0
(x2 + 4x) + (√2x + 4√2) = 0
x (x + 4) + √2 (x + 4) = 0
So x = -4, -√2 (-1.4)
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)
9. I. 6x2 – (3 + 4√3)x + 2√3 = 0,
II. 3y2 – (6 + 2√3)y + 4√3 = 0
A) x > y
B) x < y
C) x ≥ y
D) x ≤ y
E) x = y or relationship cannot be determined
Answer & Explanation Option D
Solution:
6x2 – (3 + 4√3)x + 2√3 = 0
(6x2 – 3x) – (4√3x – 2√3) = 0
3x (2x- 1) – 2√3 (2x – 1) = 0,
So x = 1/2, 2√3/3 (1.15)
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
10. I. 8x2 + (4 + 2√2)x + √2 = 0
II. y2 – (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
Answer & Explanation 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)
y2 – (3 + √3)y + 3√3 = 0
(y2 – 3y) – (√3y – 3√3) = 0
y (y – 3) – √3 (y – 3) = 0
So y = 3, √3 (1.73)