Tips and Tricks to Solve Quadratic Equation: Check Here

Tips and Tricks to Solve Quadratic Equation?

Here are the Tips and Tricks to Solve Quadratic Equation In this article, we will discuss the effective approaches to answer one of the most common questions asked is “How to solve Quadratic Equation.” Quantitative Aptitude covers one of the most important sections that are the “Quadratic Equation” in the competitive exams. It is one of the tricky and complex sections where the applicants usually make mistakes in the exam if they don’t practice it on a regular basis while preparing for the exam. As a matter of fact, there are a few tips and tricks required to Solve Quadratic Equation. With the intention to guide the students in a correct manner we have shared below few examples by which candidates can learn How to solve Quadratic Equation, Tricks to solve Quadratic Equation, and Tips on how to solve Quadratic Equation. 

 

Tricks to Solve Quadratic Equation

In bank exams, the Quadratic Equation plays an important part and is asked every year in the question paper. Generally speaking, if we observe the last year’s exam pattern then it is witnessed that 5 questions are asked from the section of Quadratic Equation every year. If the students prepare well then they can easily score these 5 marks from this section. Let’s discuss the tips and tricks to solve Quadratic Equation by following the below-discussed examples:

Quadratic Equation (Type-1)

1). Structure of a quadratic equation =X²± (Sum of Root) X ± (Product of root) = 0

QUESTION

X² – 11X + 28 = 0

Y² – 15Y + 56 = 0

 

GIVEN

In equation (i)

Sum of Root (SR) = 11

Product of Root (PR) = 28

Similarly in eq. (ii)

SR= 15

PR= 56

 

SOLUTION: NORMAL METHOD

(i). X²– 11X + 28 = 0

Now SR = -11 can be written as (-7-4 = -11)

So X² – 7X – 4X + 28 = 0

Consider the first 2 terms and take the common term outside i.e., X here

X(X – 7) – 4X + 28 = 0

Similarly consider the last 3 terms and take the common term outside i.e., -4 here

X(X – 7) – 4(X – 7) = 0

(X – 7) (X – 4) = 0

Therefore X = 7, 4

 

(ii). Y²– 15Y + 56 = 0

Now SR = -15 can be written as (-7-8 = -15)

So Y² – 7Y – 8Y + 56 = 0

Consider the first 2 terms and take the common term outside i.e., Y here

Y(Y – 7) – 8Y + 56 = 0

Similarly consider the last 3 terms and take the common term outside i.e., -8 here

Y(Y – 7) – 8(Y – 7) = 0

(Y – 7) (Y – 8) = 0

ThereforeY = 7, 8

 

So the relation between X and Y is given by both X = Y and X<Y i.e., X≤Y

 

ALTERNATE METHOD:

If the given SR is–ve then consider it as+ve

If the given SR is+ve then consider it as–ve

Split the PR into its divisible numbers such that when the numbers are added or subtracted we get the SR.

 

Here  for X value, 7 × 4 = 28 (PR) and, 7 + 4 = 11 (SR)

Here Y value,  7 × 8 = 56 (PR) and 7 + 8 = 15 (SR)

Therefore from both the equationsX = 7, 4 and Y = 7, 8

So the relation between X and Y is given by both X = Y and X<Y i.e., X≤Y

Click Here to Download the Important Concepts & Tricks for Quadratic Equation Type-1

Quadratic Equation (Type-2)

•      Structure of a quadratic equation =X²± (Sum of Root) X ± (Product of root) = 0
•      In the question discussed below the coefficient ofX²≠ 1
•    To solve these types of questions,PR (Product of root) will be taken as(PR × coefficient of X²)
•      And X = X value / coefficient of X²

QUESTION

(i)           10X²– 7X + 1 = 0

(ii)         35Y² – 12Y + 1 = 0

 

GIVEN

If the given SR is–ve then consider it as+ve

If the given SR is+ve then consider it as–ve

 

In equation (i)

Sum of Root (SR) = +7

Product of Root (PR) = 10 i.e., (1 × 10 =PR × co-efficient of X²)

 

Similarly in eq. (ii)

SR= +12

PR= 35 because (1 × 35 =PR × co-efficient of Y²)

 

SOLUTION

Split the PR into its divisible numbers such that when the numbers are added or subtracted we get the SR.

Here 5 × 2 = 10 (PR)

And 5 + 2 = 7 (SR)

 

In this type of quadratic equation, where the coefficient of X²≠ 1

X = X value / coefficient of X²

X = (5, 2) = ([5/10], [2/10]) = (0.5, 0.2)

Therefore,X = 0.5, 0.2

 

Here 7 × 5 = 35 (PR)

And 7 + 5 = 12 (SR)

Here the coefficient ofY²≠ 1

Y = Y value / coefficient of Y²

Y = (7, 5) = ([7/35], [5/35]) = (0.2, 0.14[approx.])

Therefore,Y = 0.2, 0.14

 

We have calculated the values of X and Y, now we have to compare the values with each other to deduce the relation between them

X = 0.5, 0.2; Y = 0.2, 0.14

So the relation between X and Y is given by both X = Y and X>Y i.e.,X≥Y

Click Here to Download the Important Concepts& Tricks for Quadratic Equation Type-2

Linear Equations:

The first case we will discuss is Linear Equations. In the equation, X and Y have one value. Therefore, the relation is determined in an easy manner.

Question:

6X+ 5Y=20

7X+6Y= 30

 

Solution:

(6X+ 5Y)*6=20*6,

(7X+6Y)*5= 30*5

30 X+30 Y = 120—–1

35X + 30Y = 150——-2

 

Then subtract equation 1 from equation 2, and the result will be

5X = 30

X=6, Y= 2

Hence, X>Y

Squares and Square Root Case

Question:

X= √2500

y²=1600

As the student must remember that the square root always has a positive value. Therefore X=+50 

Y= +40 AND-40. 

Hence the solution is:

+50 is greater than 40 AND-40, therefore X>Y

Squares Case: 

In this, there are both negative and positive solutions:

Question:

x²= 2500

y²=1600