Quadratic Equations for Bank Exams - Banking exam tricks

Quadratic Equations for Bank Exams 

The quadratic equation is in the form of ax² + bx +c =0 and here we need to find the value of x. Where a, b and c € R. The coefficient of X² is a Coefficient of X is b and c are constant.

Quadratic In-equations:

Let f(X) = ax² + bx +c, a≠0 and where a, b and c € R (Real numbers) be a quadratic equation, then f(x) ≥0, f(X)>0, f(X) ≤0 and f(X) <0 are known as quadratic in-equations.

These type of questions from quadratic equations can be solved by using
(1) factorization method,
(2) Formula
(3) Short Trick.

Solving these type of questions using factorization method or by the formula is little bit time-consuming. Time management is also a constraint in the success of every competitive exam. So we can solve these questions in less than 30 seconds of time by using simple speed maths trick.

Directions: In each of the following questions, two equations numbered I and II are given. You have to solve both the equations and select the appropriate option.

Give answer If
(1) x > y
(2) x = y
(3) x = y
(4) x < y
(5) Relationship between x and y cannot be established.

1. 2X² – 21X +52 =0

          2Y² -11Y + 12 =0

Factorization method: 2X^2 -8X -13X +52 = 0,
=> 2X (X-4) -13 (X-4) =0,
=> (2X-13)(X-4) = 0,
=> X=13/2, X=4.

Formula: 

Shortcut Method:
 Coefficient of X^2 is 2,
Coefficient of X is -21,
And 52 is a constant value.
Step 1: Multiply +2 and +52 i.e. +104.
Step 2: Split this +104 into two parts in such way to get there
Addition as -21 and
Multiplication value in +104.
(-13) * (-8)
Step 3: Change the sign
Step 4: Divide by 2 (Coefficient of X2).

 

2.    3X² – 13x + 14 = 0

       2Y² – 5Y + 3 = 0

Coefficient of X2 is 3,
Coefficient of X is -13
And 14 is a constant value.
Step 1: Multiply +3 and +14 i.e. +42.
Step 2: Split this +42 into two parts in such way to get there
Addition as -13 and
Multiplication value in +42.
Step 3: Change the sign
Step 4: Divide by 3 (Coefficient of X2).

3.   3X² + 7X + 2=0,
       Y² + 5Y + 6=0

4. 4X² – 8x + 3 = 0 

    4Y² –15Y + 14 = 0

4X2 – 8x + 3 = 0

Coefficient of X2 is 4,
The coefficient of X is -8.

And 3 is a constant value.
Step 1: Multiply +4 and +3 i.e. +12.
Step 2: Split this -8 into two parts in such way to get their addition as -8 and multiplication value in +12.
Step 3: Change the sign
Step 4: Divide by 4 (Coefficient of X2).
4Y² –15Y + 14 = 0
step 1: Split -15 into 2 parts in such a way that the addition of those 2 numbers should be -15 and multiplication of those 2 numbers are 56.
Step 2: Change the sign
Step 3: Divide by 4 (coefficient of X^2).

Quadratic Equation Model Questions For Practice – Previous Exam Papers

Here some problems are solved using normal factorization method but it is time-consuming in competitive exams. So solve these problems using shortcut method explained in the following video.

Quadratic Equations for Bank Exams:

In each problem two equations are given, Solve these equations to find x and y values and establish the relationship between them.

Problem 1:

I. x² – 11x + 30 = 0
II. 2y² – 9y + 10 =0

Normal Method:
(1) x² – 11x + 30 = 0
x² – 6x – 5x + 30 = 0
x (x-6) -5 (x- 6) = 0
(x-6) (x-5) = 0
=> x = 6, 5.

(2) 2y² – 9y + 10 =0
2y² – 4y – 5y + 10 =0
2y (y-2) -5 (y-2) =0
=> y = 2, 5/2.

Therefore, x > y

Problem 2:

I. 15x² + 8x + 1 =0
II. 3y² + 14y + 8 = 0

Normal Method:
(1) 15x² + 8x + 1 =0
15x² + 5x + 3x + 1 =0
5x (3x+1) + 1(3x+1)) = 0
(5x+1) (3x+1) = 0
=> x = -1/5, -1/3.

(2) 3y² + 14y + 8 = 0
3y² + 12y +2y + 8 = 0
3y (y+4) +2(y+4) =0
=> y = -4, -2/3.

Therefore, x > y

Problem 3: 
I. 4x² – 17x+ 18 = 0
II. 2y² – 21y + 40 = 0

Normal Method:
(1) 4x² – 17x+ 18 = 0
4x² -8x -9x + 18 =0
4x (x-2)-9(x-2)) = 0
(x-2)(4x-9) = 0
=> x = 2, -9/4

(2) 2y² – 21y + 40 = 0
2y² – 16y-5y + 40 = 0
2y(y-8)-5(y-8) =0
=> y = 8, 5/2.

Therefore, x < y

Problem 4: 
I. 6x² – 25x + 14 =0
II. 9y² -9y + 2 =0

Normal Method:
(1) 6x² – 25x + 14 =0
6x² – 4x -21x + 14 =0
2x (3x-2)-7(3x-2)) = 0
(2x-7)(3x-2) = 0
=> x = 7/2, 2/3

(2) 9y² -9y + 2 =0
9y² -6y -3y + 2 =0
3y(3y-2)-1(3y-2) =0
=> y = 2/3, 1/3.

Therefore, x ≥ y

Problem 5:
I.8x² + 25x + 3 =0
II. 2y² + 17y + 30 = 0

Normal Method:
(1) 8x² + 25x + 3 =0
8x² + 24x + x + 3 =0
8x (x+3)+1(x+3)) = 0
(8x+1)(x+3) = 0
=> x = -1/8, -3

(2) 2y² + 17y + 30 = 0
2y² + 12y +5y + 30 = 0
2y (y+6) +5(y+6) =0
(2y+5) (y+6) =0
=> y = -2/5, -6.

Therefore, x>y

Problem 6:
I. 3x² + 14x + 15 =0
II. 6y² + 17y + 12 = 0

Problem 7:
I. 3x²-17x +24=0
II.4y² -15y +14=0

Problem 8:
I. 2x²+11x +14=0
II. 2y² +17y +33=0

Problem 9:
I. 3x²-13x +12=0
II.2y² -15y+27=0

||. Directions: In each of the following questions, two equations are given. You
have to solve them and
a) if p < q
b) if p > q
c) if p ≤ q
d) if p ≥ q
e) if p = q

2.1) I. p2-7p = -12
II. q2-3q + 2= 0

Ans (b)

2.2) I. 12p2-7p=1
II. 6q2 – 7q + 2= 0

Ans (a)

2.3) I. p2 + 12p + 35= 0
II. 2q2 + 22q + 56= 0
Ans (c)

2.4) I. p2-8p + 15 = 0
II. q2 – 5q = -6
Ans (d)

2.5) I. 2p2 + 20p + 50 = 0
II. q2 = 25
Ans (c)

 I. 2X² +12X +16 =0

      2Y² + 14Y+24 =0.

II.  X² +13X +40 =0,

      Y² +7Y +12 =0.

III. 6X² – 7X + 2=0,

       20Y²-31Y +12 =0 

IV. 6X² +5X +1 =0,

        15Y² +8Y +1 =0

V. 88X² -19X +1 =0

     132Y² -23Y +1 =0

VI. X² +5X +6 =0,

     4Y² +24Y +35 =0.

VII. X² -24 X + 144 =0,

        Y² -26 Y +169 =0.

Ans: X<Y.

VIII.   X² + 3X -20 =0,

           2Y² +19Y + 44 =0

Ans.: X≥Y.

IX. 10X² -7X +1 =0,

        35Y² -12Y +1 =0.

Answer: X≥Y.