5 Jul 2016

Important Concepts and Tricks to Solve Quadratic Equations in Aptitude Section- Type 1 & 2 (SBI PO Special)- Download in PDF

Important Concepts and Tricks to Solve Quadratic Equations in Aptitude Section- Type 1 & 2 (SBI PO Special)- Download in PDF:
Dear Readers, we have given the Important Concepts and Tricks to Solve Quadratic Equations in Aptitude Section for SBI PO Exam 2016. Candidates those who are preparing for the examination can also download this in PDF.

QUADRATIC EQUATION (TYPE-1)

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

DIRECTIONS:

In each question below one or more equations are given on the basis of which we are supposed to find out the relationship between x and y
Give answer (1) if X>Y
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 determined

QUESTION
(i)            X2 – 11X + 28 = 0
(ii)          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
Therefore Y = 7, 8
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
Take X = 7, compare it with both the values of Y = 7, 8
We get, X = 7 is equal to Y = 7 i.e., X=Y
X = 7 is smaller than Y = 8 i.e., X<Y
Similarly Take X = 4, compare it with both the values of Y = 7, 8
We get, X = 4 is smaller than Y = 7 i.e., X<Y
X = 4 is smaller than Y = 8 i.e., X<Y
So the relation between X and Y is given by both X = Y and X<Y i.e., X≤Y

Therefore Answer is (4) if 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 7 × 4 = 28 (PR)

And 7 + 4 = 11 (SR)
Here 7 × 8 = 56 (PR)
And 7 + 8 = 15 (SR)
Therefore from both the equations X = 7, 4 and Y = 7, 8
Take X = 7, compare it with both the values of Y = 7, 8
We get, X = 7 is equal to Y = 7 i.e., X=Y
X = 7 is smaller than Y = 8 i.e., X<Y
Similarly Take X = 4, compare it with both the values of Y = 7, 8
We get, X = 4 is smaller than Y = 7 i.e., X<Y
X = 4 is smaller than Y = 8 i.e., X<Y

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

                                          QUADRATIC EQUATION (TYPE-2)


QUADRATIC EQUATION
·         Structure of a quadratic equation = X2 ± (Sum of Root) X ± (Product of root) = 0
·         In the question discussed below the coefficient of X2 ≠ 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

DIRECTIONS
In each question below one or more equations are given on the basis of which we are supposed to find out the relationship between x and y
Give answer (1) if X>Y
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 determined

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 of Y2 ≠ 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
Take X = 0.5, compare it with both the values of Y = 0.2, 0.14
We get, X = 0.5 is greater than Y = 0.2 i.e., X>Y
              X = 0.5 is greater than Y = 0.14 i.e., X>Y
Similarly Take X = 0.2, compare it with both the values of Y = 0.2, 0.14
We get, X = 0.2 is equal to Y = 0.2 i.e., X=Y
              X = 0.2 is greater than Y = 0.14 i.e., X>Y
So the relation between X and Y is given by both X = Y and X>Y i.e., X≥Y
Therefore Answer is (2) if X≥Y

0 comments:

Post a Comment

 

Receive all updates via Email

Subscribe via Email
" scrolling="no" frameborder="0" style="border:none; overflow:hidden; width:300px; height:70px;" allowtransparency="true">