# Tips & Tricks to Solve Quadratic Equations for SBI PO 2023 Exam

By Amrit Gouda|Updated : February 20th, 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 4-5 questions of Quadratic equations are asked.

You can expect around 4-5 questions from this chapter in almost all the preliminary phase of the SBI PO exam 2023. Let's discuss the tips and tricks to solve the Quadratic Equations for SBI PO exam 2023.

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 PO

• 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.
X2=1600 and Y2=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.
X2=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 X3=1331, Y3=729
then, X=11 and Y = 9
X is greater than Y, so the relation is X > Y.
If X3= -1331 and Y3= 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 X3 > Y3, the relationship is X > Y, and when X3 < Y3, the relation is X < Y.
TRICK: When both equations are in cube form. If X3 > Y3, then X > Y and X3 < Y3, then X < Y.

5. Square and cube cases.
If X2=16 and Y3=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 X2=25 and Y3=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 AX2+BX+C = 0 and AY2+BY+C = 0

 Type of Equation AX2+BX+C = 0 or AY2+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 Q-type having both roots (+).
(i)
If X2-5X+6 = 0
both roots will be positive i.e. +3 and +2
Y2-17Y+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 X2-17X+42
both roots will be positive i.e. +14 and +3.
Y2-17Y+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: Y2-49Y+444, Roots are 37,12
R type: X2+14X-1887, 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: X2-5X+6 = 0, Roots are 3,2
R type: Y2-Y-6 = 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 P-type having both roots (-) another Q-type has both roots (+).

(i) P-type: X2+5X+6=0, Roots are -3, -2
Q type: Y2-7Y+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 Q-type will always be greater than the P-type.

CASE IV: When the result of one equation is P-type having both roots negative and another is either R type or S type having one root (-) and another one (+)
(i) P-type: X2+5X+6, Roots are -3,-2
R type: X2-X-6, 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: X2+5X+6, Roots are -3, -2
R type: X2-X-6, 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 X2+X-6 = 0
Roots are -3 and +2.
Y2+5Y-66 = 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 Simplification & Approximation 2 Calculation Short Tricks 3 D.I Basics 4 Number series 5 Quadratic Equation 6 Simple Interest 7 Compound Interest 8 Mixture & Alligation 9 Partnerships 10 Percentages 11 Pipes & Cisterns 12 Age-based problems 13 Profit, Loss & Discount 14 Ratio & Proportion 15 Time, Speed & Distance 16 Time & Work 17

(ii) If X2+11X-42
Roots are -14 and +3
Y2+5Y-66 = 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 X2-X-6 = 0
Roots are +3 and -2
Y2-5Y-66 = 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 X2-11X-42
Roots are +14 and -3
Y2-5Y-66 = 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 P-type having both roots (-).
(i) If X2+5X+6 = 0
Both roots will be negative i.e. -3 and -2
Y2+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 X2+17X+42
both roots will be negative i.e. -14 and -3.
Y2+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:

2l4-36l2+162 = 0 and 3m4-75m2+432 = 0
Solution:
Basically this is not a quadratic equation because the maximum power of the variable is 4.
But if you suppose lis X and m2 is Y, then the equations will be 2X2-36X+162=0 and 3Y2-75Y+432=0
Now the converted equations are Q type having all roots positive.
X = l2 = positive roots, hence l will have 2 negative roots and 2 positive roots.
Y = m2 = positive roots, hence m will also have 2 negative roots and 2 positive roots.
So the relation between l and m cannot be defined.

Attempt quantitative aptitude quizzes from the below link -

### Quantitative Aptitude Quiz

The chapter can be useful for the following exams:

 S. No. Name of the exam 1 SBI Clerk 2 3 IBPS Clerk 4 IBPS PO 5 6 RBI Assistant 7 8 10 12

If you are aiming to crack the SBI & IBPS 2023 exam, then join the Online Classroom Program where you will get:

• Live Courses by Top Faculty
• Daily Study Plan
• Comprehensive Study Material
• Latest Pattern Test Series
• Complete Doubt Resolution
• Regular Assessments with Report Card

Why BYJU'S Exam Prep Test Series?

• Mock Tests designed by Exam Experts.
• Detailed Performance Analysis.

The most comprehensive exam prep app.

#DreamStriveSucceed

write a comment

## FAQs

• 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.

• You can expect 5 questions from it in the SBI PO exam. 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.

• You can expect around 4-5 questions from this chapter in the upcoming SBI PO Prelims 2022 exam.

• You can practice quadratic equation questions from our app or website. You need to first learn the basic concepts so that you will be able to solve the questions easily.

### PO, Clerk, SO, Insurance

BankingIBPS POIBPS ClerkSBI POIBPS SOSBI ClerkRBIIDBI SOIBPS RRBLIC