Quantitative Aptitude: Quadratic Equations Set 8 (New Pattern)

1. I. 2x2 – 15âˆš3x + 84 = 0
II. 3y2 â€“ 10âˆš3y + 9 = 0
A) x > y
B) x < y
C) x â‰¥ y
D) x â‰¤ y
E) x = y or relation cannot be established
Option A
Solution:

2x2 – 15âˆš3x + 84 = 0
Now multiply 2 and 84 = 168
we have âˆš3 in equation, so divide, 168/3 = 56
Now make factors so as by multiply you get 56, and by addition or subtraction you get â€“15
we have factors (-8) and (-7)
So 2x2 – 15âˆš3x + 84 = 0
gives
2x2 – 8âˆš3x – 7âˆš3x + 84 = 0
2x (x – 4âˆš3) – 7âˆš3 (x – 4âˆš3x) = 0
So x = 7âˆš3/2, 4âˆš3
Similarly for
3y2 â€“ 10âˆš3y + 9 = 0
Multiply 3 and 9 = 27
we have âˆš3 in equation, so divide, 27/3 = 9
Now make factors so as by multiply you get 9, and by addition or subtraction you get â€“10
we have factors (-9) and (-1)
So 3y2 â€“ 10âˆš3y + 9 = 0
gives
3y2 â€“ 9âˆš3y – âˆš3y + 9 = 0
3x (x – 3âˆš3) – âˆš3 (x – 3âˆš3x) = 0
Put all values on number line and analyze the relationship
âˆš3/3 â€¦. 3âˆš3 â€¦.. 7âˆš3/2 â€¦â€¦ 4âˆš3
2. I. x2 + âˆš5x â€“ 10 = 0
II. 2y2 + 9âˆš5y + 50 = 0
A) x > y
B) x < y
C) x â‰¥ y
D) x â‰¤ y
E) x = y or relation cannot be established
Option C
Solution:

x2 + âˆš5x â€“ 10 = 0
x2 + 2âˆš5x – âˆš5x â€“ 10 = 0
Gives x = -2âˆš5, âˆš5
2y2 + 9âˆš5y + 50 = 0
2y2 + 4âˆš5y + 5âˆš5y + 50 = 0
Gives y = -2âˆš5, -5âˆš5/2
Put all values on number line and analyze the relationship
-5âˆš5/2â€¦.. -2âˆš5â€¦.. âˆš5
3. I. 2x2 – (8+âˆš3)x + 4âˆš3 = 0
II. 3y2 â€“ (6+2âˆš3)y + 4âˆš3 = 0
A) x > y
B) x < y
C) x â‰¥ y
D) x â‰¤ y
E) x = y or relation cannot be established
Option E
Solution:

2x2 – (8+âˆš3)x + 4âˆš3 = 0
By multiplying we have to 2*4âˆš3 = 8âˆš3 and by adding/subtracting we have to get – (8+âˆš3)
So factors are -8 and -âˆš3
So 2x2 – (8+âˆš3)x + 4âˆš3 = 0
Gives
2x2 – 8x – âˆš3x + 4âˆš3 = 0
2x(x- 4) – âˆš3(x â€“ 4) = 0
So x = 4, âˆš3/2
NEXT
3y2 â€“ (6+2âˆš3)y + 4âˆš3 = 0
By multiplying we have to 3*4âˆš3 = 12âˆš3 and by adding/subtracting we have to get â€“(6+2âˆš3)
So factors are -6 and -2âˆš3
So 3y2 â€“ (6+2âˆš3)y + 4âˆš3 = 0
Gives
3y2 â€“ 6y – 2âˆš3y + 4âˆš3 = 0
3y(y- 2) – 2âˆš3(y â€“ 2) = 0
So x = 2, 2âˆš3/3
Put all values on number line and analyze the relationship
âˆš3/2â€¦â€¦ 2âˆš3/3â€¦â€¦ 2â€¦ 4
4. I. x2 – (2+âˆš5)x + 2âˆš5 = 0
II. 2y2 â€“ (6+3âˆš5)y + 9âˆš5 = 0
A) x > y
B) x < y
C) x â‰¥ y
D) x â‰¤ y
E) x = y or relation cannot be established
Option B
Solution:

x2 – (2+âˆš5)x + 2âˆš5 = 0
By multiplying we have to 2âˆš5 and by adding/subtracting we have to get – (2+âˆš5)
So factors are -2 and -âˆš5
So x2 – (2+âˆš5)x + 2âˆš5 = 0
Gives
x2 – 2x – âˆš5x + 2âˆš5 = 0
x(x- 2) – âˆš5(x â€“ 2) = 0
So x = 2, âˆš5
NEXT
2y2 â€“ (6+3âˆš5)y + 9âˆš5 = 0
By multiplying we have to 2*9âˆš5 = 18âˆš5 and by adding/subtracting we have to get â€“(6+3âˆš5)
So factors are -6 and -3âˆš5
So 2y2 â€“ (6+3âˆš5)y + 9âˆš5 = 0
Gives
2y2 â€“ 6y – 3âˆš5y + 9âˆš5 = 0
2y(y- 3) – 3âˆš5(y â€“ 3) = 0
So x = 3, 3âˆš5/2
Put all values on number line and analyze the relationship
2â€¦â€¦ âˆš5â€¦â€¦ 3â€¦ 3âˆš5/2
5. I. 3x2 + 5âˆš2x â€“ 24 = 0
II. y2 – 6âˆš2y + 16 = 0
A) x > y
B) x < y
C) x â‰¥ y
D) x â‰¤ y
E) x = y or relation cannot be established
Option B
Solution:

3x2 + 5âˆš2x â€“ 24 = 0
3x2 + 9âˆš2x – 4âˆš2x â€“ 24 = 0
Gives x = -3âˆš2, 4âˆš2/3
y2 – 6âˆš2y + 16 = 0
y2 – 2âˆš2y – 4âˆš2y + 16 = 0
Gives y = 2âˆš2, 4âˆš2
Put all values on number line and analyze the relationship
3âˆš2â€¦â€¦. 4âˆš2/3â€¦â€¦ 2âˆš2â€¦.. 4âˆš2
6. I. 3x2 – 23x + 40 = 0
II. 3y2 â€“ 8y + 4 = 0
A) x > y
B) x < y
C) x â‰¥ y
D) x â‰¤ y
E) x = y or relation cannot be established
Option A
Solution:

3x2 – 23x + 40 = 0
3x2 – 15x â€“ 8x + 40 = 0
Gives x = 5, 8/3
3y2 â€“ 8y + 4 = 0
3y2 â€“ 6y â€“ 2y + 4 = 0
Gives y = 2/3, 2
Put all values on number line and analyze the relationship
2/3â€¦.. 2â€¦.. 8/3â€¦.. 5
7. I. 5x2 â€“ 17x + 6 = 0
II. 4y2 â€“ 16y + 7 = 0
A) x > y
B) x < y
C) x â‰¥ y
D) x â‰¤ y
E) x = y or relation cannot be established
Option E
Solution:

5x2 â€“ 17x + 6 = 0
5x2 â€“ 15x â€“ 2x + 6 = 0
Gives x = 2/5, 3
4y2 â€“ 16y + 7 = 0
4y2 â€“ 2y â€“ 14y + 7 = 0
Gives y = 1/2, 7/2
Put all values on number line and analyze the relationship
2/5â€¦.. 1/2â€¦.. 3â€¦. 7/2
8. I. 3x2 – 14x + 8 = 0
II. 3y2 â€“ 20y + 12 = 0
A) x > y
B) x < y
C) x â‰¥ y
D) x â‰¤ y
E) x = y or relation cannot be established
Option E
Solution:

3x2 – 14x + 8 = 0
3x2 – 12x â€“ 2x + 8 = 0
Gives x = 4, 2/3
3y2 â€“ 20y + 12 = 0
3y2 â€“ 18y â€“ 2y + 12 = 0
Gives y = 2/3, 6
Put all values on number line and analyze the relationship
2/3â€¦â€¦.. 4â€¦.. 6
9. I. 12x2 + 25x + 12 = 0
II. 3y2 + 22y + 24 = 0
A) x > y
B) x < y
C) x â‰¥ y
D) x â‰¤ y
E) x = y or relation cannot be established
Option C
Solution:

12x2 + 25x + 12 = 0
12x2 + 16x + 9x + 12 = 0
Gives x = -4/3, -3/4
3y2 + 22y + 24 = 0
3y2 + 18y + 4y + 24 = 0
Gives y = -4/3, -6
Put all values on number line and analyze the relationship
-6â€¦â€¦ -4/3â€¦â€¦ -3/4
10. I. 6x2 + x – 2 = 0
II. 3y2 â€“ 22y + 40 = 0
A) x > y
B) x < y
C) x â‰¥ y
D) x â‰¤ y
E) x = y or relation cannot be established
Option B
Solution:

6x2 + x – 2 = 0
6x2 + 4x â€“ 3x – 2 = 0
Gives x = 1/2, -2/3
3y2 â€“ 22y + 40 = 0
3y2 â€“ 12y â€“ 10y + 40 = 0
Gives y = 10/3, 4
Put all values on number line and analyze the relationship
-2/3â€¦â€¦ 1/2â€¦â€¦ 10/3â€¦.. 4

12 Thoughts to “Quantitative Aptitude: Quadratic Equations Set 8 (New Pattern)”

1. DJ Waley Babu? ?Ritesh

Ty Maam _____________________/____________________
ðŸ™‚

2. Deepak sahu

sir exam ke point of viwe se ap dusre logo se bahut acha mattarial dete hai thank u sir

3. Deepak sahu

agar question hindi me mil jaaye to bahut aur bhi 4 star lag jayege…….

1. Sorry. Hindi me abi to available nhi ho paega

4. Shreya

5th ques me ans e ni hoga??
3âˆš2â€¦â€¦. 4âˆš2/3â€¦â€¦ 2âˆš2â€¦.. 4âˆš2
3âˆš2 2 or 4 k mid me ayega na.. scaling me

5. Hemant ahuja?

Shubra mam first five Kaise karne hai. First tyme dekhe aise ques

tell me in which step u get doubt, i ll explain

1. Hemant ahuja?

Ye squre root karne Kaise hai ques main

1. Solution me explain kiya h. Vo dekho ek bar

6. INDRAJ VERMA

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