# 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.

Guidely Online Mock Test

## 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:

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

•        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

### 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

Solution:

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

Solution:

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

Y= +40 AND-40.

Hence the solution is: +50 is greater than 40 AND-40, but -50 is smaller than +40 AND-40. Therefore the solution cannot be determined

### Cubes Cases

Question:

x3= 125

y3= 1331

Solution:

Then, X= 5, Y= 11

Y is greater than X, so the relation will beY>X.

But if in any case, the situation was, if x3= 125, y3= -1331

Then, X= 5, Y= -11

X will be greater than Y, so the relationship will beX>Y

## Tips to solve Quadratic Equation

It is very important for the students to know the basic concepts and the tricks and tips to solve the quadratic equation. They need to practice the quadratic equation on a regular basis so that they can easily and quickly attempt the questions in the actual exam. Here we have shared few tips and tricks to solve quadratic equation in the following points:

• In the first steps, the candidates need to arrange all the values on one side and leave the other side empty and then write there 0. By this, they will able to create an equation and solve it accordingly. For example, if it is given 3y2-8Y-4=3y+y2, then factor the equation as 4y2-11y-4=0 accordingly and then separate the factor set by 0.
• In the second step, the candidates need to use the formula that is −b±b2−4ac√2ab and then put the value accordingly in the equation to solve the asked questions.
• The candidates must use a stopwatch while practicing the quadratic equation as it will help them in enhancing their time management skills.
• The applicants are advised to practice daily at least 20 different types of quadratic equations to clear their basic concepts and also it will help in improving their speed and accuracy.
• The candidates must learn all the necessary short-cuts tricks to solve the quadratic equation as it will save their time in the actual exam
• They should try to attempt at least a 3-4 quadratic equation topic-wise mock test on a weekly basis. After attempting the mock test, they should analyze their strong and weak areas and then make the required improvement in those areas. 