Tips & Tricks to Solve Quadratic Equations for Bank Exams 2023
By BYJU'S Exam Prep
Updated on: September 25th, 2023
The quadratic equation is an important topic asked in the exams under the Quantitative Aptitude Section of Bank exams. In SBI PO Prelims Exam, approx 45 questions of Quadratic equations are asked.
You can expect around 45 questions from this chapter in almost all the preliminary phases of the SBI/IBPS Exams 2023. Let’s discuss the tips and tricks to solve the Quadratic Equations for SBI/IBPS Exams 2023.
Table of content
Usually, 5 questions are asked about this topic. If prepared well, you can easily score full 5 marks in this topic with the help of some short tricks. Let us discuss the important concepts of Quadratic Equations with important tricks and examples in this post now –
What is a Quadratic Equation?
An algebraic statement of the second degree in x is called a quadratic equation. A quadratic equation has the conventional form ax2 + bx + c = 0, where a and b are coefficients, x is the variable, and c is the constant term.
Importance of Quadratic Equation for SBI/IBPS Exams
 You can expect 5 questions from it in almost all the bank exams.
 These questions are very easy to solve within 2 minutes if you have practiced well.
 You can increase your chances of selection by solving these questions before approaching other questions.
First and foremost, the Relation between X and Y is established only when the relationship is defined for all solutions.
1. Linear Equations: In linear equations, both X and Y have only one value. So relations can be established easily.
4X+3Y=18, 7x+5Y= 12
(4X+3Y= 18)× 5, (7X+5Y=12)× 3
20X+15Y=90……..(i)
21X+15Y=36……..(ii)
subtracting equation (i) from equation (2)
we get, X = 54, Y = 78
Hence, Y > X
2. Squares: In this, solutions have both negative and positive values.
X^{2}=1600 and Y^{2}=3600
X = ±40 and Y = ±60
+60 is greater than both 40 and +40, but 60 is less than both 40 and +40. So, the answer will be Cannot be determined.
TRICK: Whenever both equations are given in square form, our ANSWER will be ‘Can’t be determined.’
3. Squares and Square root case.
X^{2}=1600 and Y = √3600
We know that the square root always gives a positive value. So, Y will have ONLY +60 NOT 60.
X = ±40 and Y = +60
+60 is greater than both +40 and 40. Hence Y > X.
4. Cubes Case.
If X^{3}=1331, Y^{3}=729
then, X=11 and Y = 9
X is greater than Y, so the relation is X > Y.
If X^{3}= 1331 and Y^{3}= 729
then, X= 11 and Y = 9
X is greater than Y, so the relation is X < Y.
Note: Can you see something common in the above example? Common thing is that when X^{3 }> Y^{3}, the relationship is X > Y, and when X^{3} < Y^{3}, the relation is X < Y.
TRICK: When both equations are in cube form. If X^{3 }> Y^{3}, then X > Y and X^{3 }< Y^{3}, then X < Y.
5. Square and cube cases.
If X^{2}=16 and Y^{3}=64
then X = +4, 4 and Y=4
So, Y = 4 is equal to X = 4 and Y = 4 is greater than X = 4.
So, Y ≥ X
If X^{2}=25 and Y^{3}=64
then X = +5, 5 and Y = 4
So, Y = 4 is greater than X = 5 and less than X = +5, So the relation Can’t be Determined.
Table Method to Solve Quadratic Equations Easily
1. Write down the table (given below) before the exam starts, in your rough sheet, to use during the exam, Analyse the (+, ) signs in the problem, and refer to the table of signs.
2. Write down the new (solution) signs, and see if a solution is obtained instantly. If not, then go to step 3.
3. Obtain the two possible values for X & Y, from both equations,
4. Rank the values and get the solution,
STEP 1
Firstly, when you enter the exam hall, you need to write down the following master table in your rough sheet instantly (only the signs):
Let us consider that the equations are AX^{2}+BX+C = 0 and AY^{2}+BY+C = 0
Type of Equation

AX^{2}+BX+C = 0 or AY^{2}+BY+C = 0 
Roots in X or Y equation 

Sign of BX or BY 
Sign of C 
Sign of bigger root 
Sign of smaller root 

P 
+ 
+ 
– 
– 
Q 
– 
+ 
+ 
+ 
R 
+ 
– 
– 
+ 
S 
– 
– 
+ 
– 
Now we will discuss the cases as mentioned below in the table.
CASE 
ROOTS OF X /Y 
ROOTS OF X/Y 
CONCLUSION 
I 
+,+ (Q) 
+,+ (Q) 
Easy 
II 
+,+ (Q) 
+, (R or S) 
Will discuss 
III 
+,+ (Q) 
, (P) 
Left>Right 
IV 
+, (R or S) 
, (P) 
Will discuss 
V 
+, (R or S) 
+, (R or S) 
Cannot be defined 
VI 
, (P) 
, (P) 
Easy 
CASE I: When the result of both equations are Qtype having both roots (+).
(i) If X^{2}5X+6 = 0
both roots will be positive i.e. +3 and +2
Y^{2}17Y+66 = 0
both roots will be positive i.e. +11 and +6
We can see that both roots of X are less than both roots of Y. So, X < Y.
(ii) If X^{2}17X+42
both roots will be positive i.e. +14 and +3.
Y^{2}17Y+66 = 0
both roots will be positive i.e. +11 and +6.
Here, +14 > +11
but +14 < +6
also +3 < +11
and +3 < +6
As we can see in the comparison above, there are TWO relations between X and Y which are both > and <. So relation cannot be defined.
Note 1: When both equations have BX () and C(+), You have to go into detail.
CASE II: When the result of one equation is Q type and another is either R type or S type.
(i) Q type: Y^{2}49Y+444, Roots are 37,12
R type: X^{2}+14X1887, Roots are 51,37
Now let us compare the values in the table below –
X 
RELATION 
Y 
51 
< 
37 
51 
< 
12 
37 
= 
37 
37 
> 
12 
When we compared the values of X and Y in the table above, we found that there are THREE relations between X and Y i.e. =, > and <. So, a relation cannot be defined.
(ii) Q type: X^{2}5X+6 = 0, Roots are 3,2
R type: Y^{2}Y6 = 0, Roots are 3,2
Now let us compare the values in the table below –
X 
RELATION 
Y 
3 
= 
3 
3 
> 
2 
2 
< 
3 
2 
> 
2 
When we compared the values of X and Y in the table above, we found that there are TWO relations between X and Y i.e. >, =. So the relation CANNOT BE DEFINED.
CASE III: When one equation is Ptype having both roots () another Qtype has both roots (+).
(i) Ptype: X^{2}+5X+6=0, Roots are 3, 2
Q type: Y^{2}7Y+12=0, Roots are 4,3
Comparing the values in the below table –
X 
RELATION 
Y 
3 
< 
3 
3 
< 
2 
2 
< 
3 
2 
< 
2 
On comparing, we saw that the So roots of the Y equation are greater than the roots of X.
Note: In this case, the roots of the equation Qtype will always be greater than the Ptype.
CASE IV: When the result of one equation is Ptype having both roots negative and another is either R type or S type having one root () and another one (+)
(i) Ptype: X^{2}+5X+6, Roots are 3,2
R type: X^{2}X6, roots are 3,2
Comparing the values in the below table –
X 
RELATION 
Y 
3 
< 
3 
3 
< 
2 
2 
< 
3 
2 
= 
2 
In a comparison of X and Y values, There are THREE relations between X and Y i.e. =, > and <. So relation cannot be defined.
(ii) P type: X^{2}+5X+6, Roots are 3, 2
R type: X^{2}X6, Roots are 3,2
Now let us compare the values in the table below –
X 
RELATION 
Y 
3 
< 
3 
3 
< 
2 
2 
< 
3 
2 
= 
2 
When we compared the values of X and Y in the table above, we found that there are TWO relations between X and Y i.e. <, =. So, the relation is X ≤ Y.
Case V: When the result of both equations is either R type or S type or one equation is R type and another is S type having one root () and another root (+).
(i) If X^{2}+X6 = 0
Roots are 3 and +2.
Y^{2}+5Y66 = 0
Roots are 11 and +6.
Comparing the values in the below table –
X 
RELATION 
Y 
3 
> 
11 
3 
< 
+6 
+2 
> 
11 
+2 
< 
+6 
In the comparison of X and Y values, there are two relations between X and Y i.e. both > and <. So, the relation cannot be defined.
You can check other quantitative Aptitude study materials here:
S.No 
Topics 
1 

2  
3 
D.I Basics 
4 

5 

6 

7 

8 

9 

10 

11 

12 

13 

14 

15 

16 

17 
(ii) If X^{2}+11X42
Roots are 14 and +3
Y^{2}+5Y66 = 0
Roots are 11 and +6
Comparing the values in the below table –
X 
RELATION 
Y 
14 
< 
11 
14 
< 
+6 
+3 
> 
11 
+3 
< 
+6 
When we compared the values of X and Y in the table above, we saw that there are two relations between X and Y i.e. both > and <. So, the relation cannot be defined.
(iii) If X^{2}X6 = 0
Roots are +3 and 2
Y^{2}5Y66 = 0
Roots are +11 and 6.
Let us compare the values of X and Y the below table –
X 
RELATION 
Y 
+3 
< 
+11 
+3 
> 
6 
2 
< 
+11 
2 
> 
6 
As the table shows, there are two relations between X and Y i.e. both > and <. So, the relation cannot be defined.
(iv) If X^{2}11X42
Roots are +14 and 3
Y^{2}5Y66 = 0
Roots are +11 and 6
We are comparing the roots of X and Y in the table below –
X 
RELATION 
Y 
+14 
> 
+11 
+14 
> 
6 
3 
< 
+11 
3 
> 
6 
On comparing, there are two relations between X and Y i.e. both > and <. So relation cannot be defined.
Note 4: In this case, the answer will always be CANNOT BE DEFINED.
CONCLUSION: Whenever in a question, the sign of C is negative () in both the X and Y equation, then the answer will always be CANNOT BE DEFINED.
Case VI: When the result of both equations is Ptype having both roots ().
(i) If X^{2}+5X+6 = 0
Both roots will be negative i.e. 3 and 2
Y^{2}+17Y+66 = 0
Both roots will be negative i.e. 11 and 6
We can see that both roots of X are greater than both roots of Y. So, X > Y.
(ii) If X^{2}+17X+42
both roots will be negative i.e. 14 and 3.
Y^{2}+17Y+66 = 0
both roots will be negative i.e. 11 and 6.
On comparing these values of X and Y in the table below –
X 
RELATION 
Y 
14 
< 
11 
14 
< 
6 
3 
> 
11 
3 
> 
6 
We found that there are two relations between X and Y i.e. both > and <. So relation cannot be defined.
Note 6: When both equations have BX () and C(+), You have to go into detail.
Other Examples:
2l^{4}36l^{2}+162 = 0 and 3m^{4}75m^{2}+432 = 0
Solution: Basically this is not a quadratic equation because the maximum power of the variable is 4.
But if you suppose l^{2 }is X and m^{2} is Y, then the equations will be 2X^{2}36X+162=0 and 3Y^{2}75Y+432=0
Now the converted equations are Q type having all roots positive.
X = l^{2} = positive roots, hence l will have 2 negative roots and 2 positive roots.
Y = m^{2} = positive roots, hence m will also have 2 negative roots and 2 positive roots.
So the relation between l and m cannot be defined.
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