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# Quadratic Equation Shortcut Tricks For Bank Exams –  Speed Maths Tricks

Learning Quadratic Equation Shortcut Tricks for bank exams is important for those who are preparing for IBPS Bank Exams,  SBI Exams, SSC, Railways, Insurance, etc.,

The quadratic equation is in the form of ax2 + bx +c =0 and here we need to find the value of x. According to the analysis of previous bank exam papers, we can say that 5 questions are asked of quadratic equations.

Here we are sharing the simple speed maths trick by which we can solve quadratic equations in quick time instead of using traditional methods like factorization and formula, use Quadratic Equation Shortcut Tricks For Bank Exams to solve x and y values using a shortcut to compare two quadratic equations.

In IBPS Bank Exams, We have a Quantitative Aptitude section in that 50 questions are asked from various topics of QA (Quantitative Aptitude).

## Quadratic Equation Shortcuts for Bank Exams

Here, we are explaining the traditional methods and shortcut methods to solve Quadratic Equation problems in step by step methods.

#### In the following questions, two equations numbered I and II are given. You have to solve both the equations and

Give answer (2) if x ≥ y
Give answer (3) if x < y
Give answer (4) if x ≤y
Give answer (5) if x=y (Or) the relationship cannot be established.

Problem 1

#### II. 5y2 – 51y + 54 =0.

Normal Method: (Factorization Method)

I. 40x2 – 47x +12 =0

40x2 – 32x-15x +12 =0

8x(5x– 4) – 3(5x -4) =0
(8x-3) (5x-4)=0
8x-3 =0, 5x-4 =0
X= 3/8,  4/5.

II. 5y2 – 51y + 54 =0.
5y2 – 45y -6y + 54 =0.
5y (y-9) -6(y-9)=0,
(5y-6 ) (y-9) =0
6/5, 9
Clearly, x < y  Answer (1)

Quadratic Equation Shortcut Trick to solve :40x2 – 47x +12 =0

1) split the middle term (X coefficient). so as to get the product of two numbers should be (12 *40) = 480,
2) the sum of the two numbers is -47 and

3) Change the sign of split numbers i.e -32, -15 becomes +32, +15.

4) Divide +32 and +15 with X^2 coefficient 1.e. 40

5) Then we get x = 4/5, 3/8.

Apply the same procedure to second equation and find y value.

Find the relationship between x and y values to compare two quadratic equations. Clearly, x < y  Answer (1)

#### II. 63y2 – 11y -40 =0.

Factorization Method:

I. 22x2 – x -6 =0.

• 22x2 – 12x + 11x -6 =0.
• 11x( 2x+1) – 6(2x+1) =0
• (11x-6)(2x+1)=0

X=6/11, -1/2.

II. 63y2 – 11y -40 =0.

• 63y2 – 56y + 45y – 40 =0.
• 7y (9y-8) + 5(9y -8) =0,
• (7y+5) (9y-8) =0,

y = -5/7,  8/9.

Here we have to compare two fractions to know the relation between x and y that means x=y or x> y or x>y. Shortcut Trick to compare two fractions #### Quadratic Equation Short Trick Explanation:

1) Split the middle term (X coefficient) i.e. (-1) in such a way to get the product of two numbers should be (22 * (-6)) = 132,

2) the sum of the two numbers is -1 and

3) Change the sign of split numbers i.e -12, +11 becomes +12, -11.

4) Divide +12 and -11 with X^2 coefficient 1.e. 22

5) Then we get x = 6/11, 1/2.

Apply the same procedure to second equation and find y value observe the above image by reading the explanation you can understand the shortcut trick.

Find the relationship between x and y values to compare two quadratic equations.

#### I. 20x2 – 37x + 8 =0.

• 20x2 – 32x -5x + 8 =0.
• 4x(5x-8)-1(5x-8)=0,
• (4x-1)(5x-8)=0,

= ¼, 8/5.

II. 24y2 +38y -7 =0.

• 24y2 +42y-4y -7 =0.
• 6y (4y +7)-1(4y+7)=0
• (4y+7)(6y-1)=0

= -7/4 or 1/6. #### II. 4y2 – 20y + 21 =0.

12x2 – 28x +15 =0.

• 12x2 – 10x -18x +15 =0.
• 2x (6x – 5) -3(6x -5) =0.
• (2x-3) (6x-5) =0

= 2/3, 5/6.

4y2 – 20y + 21 =0.

• 4y2 – 14y-6y + 21 =0.
• 2y( 2y-7) – 3(2y-7) =0

= 3/2, 7/2. #### II. 27y2 – 12y + 1 =0. Clearly x > y. Answer (1).

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 a 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 Quadratic Equation Short Tricks.

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.

(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.

The coefficient of X^2 is 2,
A 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
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 The coefficient of X2 is 3,
The 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
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 #### 4Y2 –15Y + 14 = 0

4X2 – 8x + 3 = 0
Coefficient of X2 is 4,
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).

## Use the shortcut to compare two quadratic equations

Here some problems are solved using normal factorization method but it is time-consuming in competitive exams. To 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 by using Quadratic Equation Shortcut Tricks to compare two equations and then if we get x and y value in fractions then use shortcut method for comparing two fractions.

### Quadratic formula examples step by step

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 the constant.

Formula: Quadratic Equations solving by using either Formula or using factorization method is the traditional solving methods in maths but using Quadratic Equation Shortcut Tricks For Bank Exams, we get answers quickly.

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.

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