**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-**

- I. 3x
^{2} + 10x – 8 = 0,

II. 2y^{2} – 13y + 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:

3x^{2} + 10x – 8 = 0

3x^{2} + 12x – 2x – 8 = 0

Gives x = -2, 2/3

2y^{2} – 13y + 6 = 0

2y^{2} – 12y – y + 6 = 0

Gives y = 1/2, 6

- I. 16x
^{2} + 8x – 15 = 0,

II. 4y^{2} + 29y + 30 = 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:

16x^{2} + 8x – 15 = 0

16x^{2} + 20x – 12x – 15 = 0

Gives x = -5/4, 3/4

4y^{2} + 29y + 30 = 0

4y^{2} + 24y + 5y + 30 = 0

Gives y = -6, -5/4

- I. 3x
^{2} – 25x + 52 = 0,

II. 15y^{2} – 38y – 40 = 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:

3x^{2} – 25x + 52 = 0

3x^{2} – 12x – 13x + 52 = 0

Gives x = 4, 13/3

15y^{2} – 38y – 40 = 0

15y^{2} + 12y – 50y – 40 = 0

Gives y = -4/5, 10/3

- I. 12x
^{2} – 5x – 3 = 0,

II. 4y^{2} – 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 D**

Solution:

12x^{2} – 5x – 3 = 0

12x^{2} + 4x – 9x – 3 = 0

Gives x = -1/3, 3/4

4y^{2} – 11y + 6 = 0

4y^{2} – 8y – 3y + 6 = 0

Gives y= 3/4, 2

- I. 3x
^{2} + 7x – 6 = 0,

II. 6y^{2} – y – 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 E**

Solution:

Explanation:

3x^{2} + 7x – 6 = 0

3x^{2} + 9x – 2x – 6 = 0

Gives x = -3, 2/3

6y^{2} – y – 2 = 0

6y^{2} + 3y – 4y – 2 = 0

Gives y = -1/2, 2/3

Put on number line

-3… -1/2 … 2/3

- I. 4x
^{2} + 15x + 9 = 0,

II. 4y^{2} – 13y – 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:

4x^{2} + 15x + 9 = 0

4x^{2} + 12x + 3x + 9 = 0

Gives x = -3, -3/4

4y^{2} – 13y – 12 = 0

4y^{2} – 16y + 3y – 12 = 0

Gives y = -3/4, 4

- I. 2x
^{2} – (6 + √3)x + 3√3 = 0,

II. 3y^{2} – (9 + √3)y + 3√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 E**

Solution:

2x^{2} – 6x – √3x + 3√3 = 0

2x (x- 3) – √3 (x – 3) = 0,

So x = 3, √3/2 (0.7)

3y^{2} – 9y – √3y + 3√3 = 0

3y (y – 3) – √3 (y – 3) = 0

So x = 3, √3/3 (0.6)

- I. 2x
^{2} – (2 + 2√5)x + 2√5 = 0

II. 4y^{2} – (6 + 2√2)y + 3√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 E**

Solution:

2x^{2} – 2x – 2√5x + 2√5 = 0

2x (x – 1) – 2√5 (x – 1) = 0

So x = 1, √5 (2.2)

4y^{2} – 6y – 2√2y + 3√2 = 0

2y (2y – 3) – √2 (2y – 3) = 0

So, y = 3/2, 1/√2 (0.7)

- I. 2x
^{2} – 15√3x + 84 = 0,

II. 3y^{2} – 2y – 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 A**

Solution:

2x^{2} – 15√3x + 84 = 0

Now multiply 2 and 84 = 168

we have √3 in equation, so divide, 168/3 = 56

Now make factors so as by multiply you get 56, and by addition or subtraction you get –15

we have factors (-8) and (-7)

So 2x^{2} – 15√3x + 84 = 0

gives

2x^{2} – 8√3x – 7√3x + 84 = 0

2x (x – 4√3) – 7√3 (x – 4√3x) = 0

So x = 3.5√3, 4√3

3y^{2} – 2y – 8 = 0

3y^{2} – 6y + 4y – 8 = 0

So y = -4/3, 1

Plot on number line

-4/3……1….. 3.5√3….. 4√3

- I. 16x
^{2} + 20x + 6 = 0

II. 10y^{2} + 38y + 24 = 0

A) x > y

B) x < y

C) x ≥ y

D) x ≤ y

E) x = y or relationship cannot be determined

View Answer

** Option A**

Solution:

Divide both equations by 2

8x^{2} + 10x + 3 = 0

8x^{2} + 4x + 6x + 3 = 0

Gives x = -1/2, -3/4

5y^{2} + 19y + 12 = 0

5y^{2} + 15y + 4y + 12 = 0

Gives y = -4, -4/5

ty mam 🙂

@Shubhra_AspirantsZone:disqus mam…… in 10th ques ans shldn’t be A???

yes A

ty mam 🙂

10?? th how

check solution.

ok mam

10th?????????????????????//

check solution

I. 2×2 – (2 + 2√5)x + 2√5 = 0

II. 4y2 – (6 + 2√2)y + 3√2 = 0

MAAM AISE QUESTIONS KO SIMPLIFY KARNA PADEGA USKE BAAD HI SOLVE KARNA HOGA

YA KOI SHORT TRICK HAI SOLVE KARNE KA?

U know the formula of discriminant method?

NO MAAM BATA DJIYE

SHAYAD PATA HO PAR NAAM SUNKE YAAD NHI AARHA

MAAM Q-9 TYPE K AUR QSNS HAI?

Check here

http://aspirantszone.com/quantitative-aptitude-quadratic-equations-set-8-new-pattern/

I ll add more

OKAY MAAM

THANK YOU VERY MUCH:)

ty mam:))

nice ques mam

thanku

Thank you mam……….

tyy mam:)

ty..

ty mam…. 7 n 8 are easy to solve but one has to noe about roots till 10

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