Quadratic Equations Set – 24

1. I. 5x2  – 18x + 9 =0
II. 3y2 + 5y – 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 A
Solution:

5x2   – 18x + 9  = 0
=> 5x2  – 15x – 3x  + 9 =0
=> (5x – 3 )(x-3 )= 0
=> x=  3/ 5 or  x= 3
3y2 + 5y – 2 = 0
=> 3y2  + 6y – y -2 = 0
=> (3y-1)(y + 2) = 0
=> y = 1/3  or  -2

 

2. I. √x  – √6 / √x = 0
II. y3 – 63/2  = 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: 

√x  –   √6 / √x = 0
x – √6 = 0
x = √6
y3   – 6 (3/2 )  = 0
=>y3  = (√6)3
=> y = √6

 

3. I. (625)1/4x + √1225  = 155
II. √196y + 13  = 279
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: 

5x + 35 = 155
=> 5x = 155 – 35
=> x = 120/ 5 = 24√196 y  + 13 = 279
=> 14y = 279 -13
=> y = 266/14 =19

 

4. I. 3x2 – 17x + 24 =0
II. 4y2  – 15y + 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: 

3x2  -17x  + 24 =0
=> 3x2 – 9x  – 8x +24 = 0
=> (3x- 8)(x-3) = 0
=>  x=  8/3 or 3
4y2  – 15y + 14 = 0
=> 4y2  – 8y -7y +14 = 0
=> (4y -7 )(y-2)=0
=> y = 7 /4 or  2
 

 

5. I. x2 – 2x – √5 x + 2√5 = 0
II. y2 – √3 y – √2 y + √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 A
Solution: 

x2  – 2x- √5 x + 2√5 = 0
=> x(x-2 ) – √5 (x-2 )= 0
=> (x-2)(x-√5)=0
=> x= 2 0r √5
y2  – √3 y  – √2 y + √6 =0
=> y(y-√3) – √2(y – √3) = 0
=> (y – √2)(y- √3) =0
=> y = √2 or √3

 

6. I. 5x2 – 23x + 12 = 0,
II. 5y2 – 28y + 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

Answer & Explanation

  Option E
Solution: 

x = 3/5, 4
y = 3/5, 5

 

7. I. 6x2 + 5x – 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

Answer & Explanation

 Option D
Solution: 

x = 2/3, 3
y = -3/2, 2/3

 

8. I. 3x2 – 5x – 12 = 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 B
Solution: 

x = -4/3, 3
y = 4, 9/2

 

9. I. 8x2 – (4 + 4√3)x + 2√3 = 0,
II. 3y2 – (4 + 3√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 B
Solution: 

8x2 – (4 + 4√3)x + 2√3 = 0
(8x2 – 4x) – (4√3x – 2√3) = 0
4x (2x- 1) – 2√3 (2x – 1) = 0,
So x = 1/2 (0.5), 2√3/4 (0.87)
3y2 – (4 + 3√3)y + 4√3 = 0
(3y2 – 4y) – (3√3y – 4√3) = 0
y (3y – 4) – √3 (3y – 4) = 0
So, y = √3 (1.732), 4/3
put on number line
0.5………..0.87……..4/3………1.732

 

10. I. x2 + (4 + 2√2)x + 8√2 = 0
II. 3y2 – (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√2)x + 8√2 = 0
(x2 + 4x) + (2√2x + 8√2) = 0
x (x + 4) + 2√2 (x + 4) = 0
So x = -4, -2√2 (-2.8)
3y2 – (3 + √3)y + √3 = 0
(3y2 – 3y) – (√3y – √3) = 0
3y (y – 1) – √3 (y – 1) = 0
So y = 1, √3/3 (0.57)

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