300+ Most Asked Quadratic Equation for Bank Exam

Navigating the waters of bank exams can be challenging, and one topic that continually surfaces is the quadratic equation for bank exam. This area has consistently been of significant importance, with the quadratic equation for bank exam featuring prominently in many papers. Notably, questions from this topic, especially in the first phase, come mainly in the form of inequalities. However, the breadth of the quadratic equation for bank exam extends beyond that, assisting in solving various word problems by forming equations.

As many candidates gear up for the bank exams of 2023, they are actively seeking quadratic equation questions for bank exams, quadratic equation questions for banking pdf, and quadratic equation questions for bank po to bolster their preparation. These resources are invaluable for those looking to develop a strong command over the quadratic equation for bank topic and aim to answer efficiently without consuming too much time.

However, mastering the quadratic equation for bank exam isn’t just about understanding the concept. It’s also about knowing the right techniques. That’s where quadratic equation tricks for bank exam come into play, offering candidates strategic ways to tackle these problems. As the demand for quadratic equation questions for bank exams, quadratic equation questions for banking pdf, and quadratic equation questions for bank po rises, it’s clear that a comprehensive grasp on quadratic equation for bank and its nuances can make all the difference.

Directions:1-10) In each of the following questions two equations Ⅰ and Ⅱ are given. You have to solve both the equations and give answer

9P²– 9P + 2 = 0
18Q²+ 3Q = 1

P < Q
P ≤ Q
P ≥ Q
P = Q or no relation can be established between P and Q
P > Q

Ans:5
Ⅰ. 9P² – 9P + 2 = 0

Or, 9P² – 6P – 3P + 2 = 0

Or, (3P – 1)(3P – 2) = 0

P = (1 / 3), (2 / 3)

Ⅱ. 18Q² + 3Q – 1 = 0

Or, 18Q² + 6Q – 3Q – 1 = 0

Or, (6Q – 1)(3Q + 1) = 0

Q = – (1 / 3), (1 / 6)

Hence, P > Q

I. P²+ 13P + 42 = 0
2Q²+ 22Q + 60 = 0

P < Q
P ≤ Q
P ≥ Q
P = Q or no relation can be established between P and Q
P > Q

Ans:2
Ⅰ. P² + 13P + 42 = 0

Or, P² + 7P + 6P + 42 = 0

Or, (P + 7)(P + 6) = 0

∴ P = -7, -6

Ⅱ. 2Q² + 22Q + 60 = 0

Or, Q² + 11Q + 30 = 0

Or, (Q + 5)(Q + 6) = 0

∴ Q = -5, -6

Hence , Q ≥ P

I. 6P + 7Q = 45
2P + 3Q = 17

P < Q
P ≤ Q
P ≥ Q
P = Q or no relation can be established between P and Q
P > Q

Ans:5
P = 4 and Q = 3

Hence P > Q

I. 3P²+ 192 = – 48P
Q²+ 16Q + 64 = 0

P < Q
P ≤ Q
P ≥ Q
P = Q or no relation can be established between P and Q
P > Q

Ans:4
Ⅰ. 3P² + 48P + 192 = 0

Or, P² + 16P + 64 = 0

Or, (P + 8)² = 0

∴ P = -8

Ⅱ. Q² + 16Q + 64 = 0

Or, (Q + 8)² = 0

∴ Q = -8

Hence , P = Q

I. 15P²– 8P + 1 = 0

II. 45Q²+ 21Q = 6

P < Q
P ≤ Q
P ≥ Q
P = Q or no relation can be established between P and Q
P > Q

Ans:3
Ⅰ. 15P² – 8P + 1 = 0

Or, 15P² – 3P – 5P + 1 = 0

Or, (3P – 1)(5P – 1) = 0

∴ P = (1 / 3), (1 / 5)

Ⅱ. 45Q² + 21Q – 6 = 0

15Q² + 7Q – 2 = 0

Or, 15Q² + 10Q – 3Q – 2 = 0

Or, (5Q – 1)(3Q + 2) = 0

∴ Q = (1 / 5), – (2 / 3)

Hence, P ≥ Q

I. X – 8/√X=0
II. √Y- √7/√Y=0

If X > Y
If X ≥ Y
If X< Y
If X≤ Y
X=Y or Relationship cannot be established.

Ans:1
I. X – 8/√X=0

Or, x^3/2= 8

Or, X = 4

II. √Y- √7/√Y=0

Or, Y = √7

Y = 2.6

Hence X > Y

I. 8X²+78X+169=0

II. 5Y²+ 14Y+8=0

If X > Y
If X ≥ Y
If X< Y
If X≤ Y
X=Y or Relationship cannot be established.

Ans:3
I. 8X²+78X+169=0

Or, 8X²+52X+26X+169=0

Or, 4X(2X+13)+ 13(2X+13)=0

Or, (4X+13)(2X+13)=0

X= (-13)/4,(-13)/2

II.5Y²+ 14Y+8=0

Or, 5Y²+10Y+4Y+8=0

Or, 5Y(Y+2)+4(Y+2)=0

Or, (5Y+4)(Y+2)=0

Y= -2 , (-4)/5

Hence X< Y

Pages: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30