Quadratic Equations Set – 17

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

Answer & Explanation

 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. 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

Answer & Explanation

 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. 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

Answer & Explanation

 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, 1/2

 

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

Answer & Explanation

 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. 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

Answer & Explanation

 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. 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

Answer & Explanation

 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. 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

Answer & Explanation

 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. 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

Answer & Explanation

 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. 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

Answer & Explanation

 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. 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

Answer & Explanation

 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

Leave a Comment

Your email address will not be published.