** 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 =

**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 = 0Now SR = -11 can be written as (-7-4 = -11)

So X

^{2}– 7X – 4X + 28 = 0Consider 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 = 0Now SR = -15 can be written as (-7-8 = -15)

So Y

^{2}– 7Y – 8Y + 56 = 0Consider 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, 8We 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, 8We 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, 8We 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, 8We 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, 0.14We 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, 0.14We 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