Quantitative Aptitude: Quadratic Equations Set 22

Quadratic Equations Practice Sets for IBPS PO, NICL, NIACL, 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-

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
   
   View Answer

    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
   
   View Answer

    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
   
   View Answer

    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
   
   View Answer

    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
   
   View Answer

    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
   
   View Answer

    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
   
   View Answer

   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
   
   View Answer

   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
   
   View Answer

   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
    
    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)
    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)
    

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