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__Important Concepts and Tricks to Solve Quadratic Equations in Aptitude Section- Type 1 & 2 (SBI PO Special)- Download in PDF__
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Tricks to Solve Quadratic Equations in Aptitude Section for SBI PO Exam 2016. Candidates those who are preparing for the examination can also
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__QUADRATIC EQUATION (TYPE-1)__

1). Structure of
a quadratic equation =

**X**^{2 }± (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)
X

^{2}– 11X + 28 = 0
(ii)
Y

^{2}– 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). X

^{2}– 11X + 28 = 0
Now SR = -11 can be written as (-7-4 = -11)

So X

^{2}– 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). Y

^{2}– 15Y + 56 = 0
Now SR = -15 can be written as (-7-8 = -15)

So Y

^{2}– 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
=

**X**^{2 }± (Sum of Root) X ± (Product of root) = 0
·
In the question discussed below
the coefficient of

**X**^{2 }≠ 1
·
To solve these types of questions,

**PR (Product of root)**will be taken as**(PR × coefficient of X**^{2})
·
And

**X = X value / coefficient of X**^{2}

^{}__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)
10X

^{2 }– 7X + 1 = 0
(ii)
35Y

^{2}– 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 X**)

^{2}
Similarly in eq. (ii)

**SR**= +12

**PR**= 35 because (1 × 35 =

**PR × co-efficient of Y**)

^{2}__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

**X**^{2 }≠ 1**X = X value / coefficient of X**

^{2}
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

**Y**^{2 }≠ 1**Y = Y value / coefficient of Y**

^{2}
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

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