Quantitative Aptitude: Quadratic Equations 

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-

1. I. 3x2 – 25x + 52 = 0,
   II. 3y2 – 16y + 16 = 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
   
2. I. 3x2 – 8x – 16 = 0,
   II. 2y2 – 5y – 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
   
3. I. 3x2 + 22x + 24 = 0,
   II. 2y2 + 7y – 4 = 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
   
4. I. 2x2 – 9x + 4 = 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
   
5. I. 3x2 + 7x – 6 = 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
   
6. I. 3x2 – 4x – 4 = 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
   
7. I. 7x2 + 19x – 6 = 0,
   II. 2y2 + 13y + 21 = 0
   A) x > y
   B) x < y
   C) x ≥ y
   D) x ≤ y
   E) x = y or relationship cannot be determined
   
8. I. x2 + (4 + √2)x + 4√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
   
9. 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
   
10. I. 4x2 – 12x + 5 = 0,
    II. 2y2 + 3y – 20 = 0
    A) x > y
    B) x < y
    C) x ≥ y
    D) x ≤ y
    E) x = y or relation cannot be established

SOLUTION 

1.      Option C
Solution: 
3x² – 25x + 52 = 0
3x² – 12x – 13x + 52 = 0
Gives x = 4, 13/3
3y² – 16y + 16 = 0
3y² – 14y – 4y + 16 = 0
Gives y = 4, 4/3

2.      Option E
Solution: 
3x² – 8x – 16 = 0
3x² – 12x + 4x – 16 = 0
Gives x = -4/3, 4
2y² – 5y – 12 = 0
2y² – 8y + 3y – 12 = 0
Gives y = -3/2, 4

3.      Option E
Solution: 
3x² + 22 x + 24 = 0
3x² + 18x + 4x + 24 = 0
So x = -4/3, -6
2y² + 7y – 4 = 0
2y² + 8y – y – 4 = 0
Gives y = -4, ½

4.      Option D
Solution: 
2x² – 9x + 4 = 0
2x² – 8x – x + 4 = 0
So x = 4 , 1/2
2y² – 17y + 36 = 0
2y² – 8y – 9y + 36 = 0
Gives y= 4, 9/2

5.      Option B
Solution: 
Explanation:
3x² + 7x – 6 = 0
3x² + 9x – 2x – 6 = 0
Gives x = -3, 2/3
3y² – 19y + 20 = 0
3y² – 15y – 4y + 20 = 0
Gives y = 4/3, 5

6.      Option A
Solution: 
3x² – 4x – 4 = 0
3x² – 6x + 2x – 4 = 0
Gives x = -2/3, 2
3y² + 16y + 20 = 0
3y² + 6y + 10y + 20 = 0
Gives y = -10/3, -2

7.      Option C
Solution: 
7x² + 19x – 6 = 0
7x² + 21x – 2x – 6 = 0
Gives x = -3, 2/7
2y² + 13y + 21 = 0
2y² + 6y + 7y + 21 = 0
So y = -7/2, -3

8.      Option B
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)
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)

9.      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
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, -√2

10. Option E
Solution: 
4x² – 12x + 5 = 0
4x² – 2x – 10x + 5 = 0
x = 1/2, 5/2
2y² + 3y – 20 = 0
2y² + 8y – 5y – 20 = 0
So y = -3, 5/2