LET US SEE HOW TO SOLVE QUADRATIC EQUATION-HOW TO SERIES

Hi Bankersdaily Aspirants,

Aspirants today we are going to discuss on the most important topic in Aptitude Section that is Quadratic Equation.Quadratic Equation is asked for 5 marks in Prelims & Mains Examination.This is the easiest section where we can score 5 marks in less than 5 minutes.In this Article we are going to discuss some of the pattern following which Question on Quadratic Equation  is asked.

TOPICS TO BE DISCUSSED

1)BASED ON BASIC QUADRATIC EQUATION

2)BASED ON ROOTS

3)BASED ON FINDING VALUE

#.1.TYPE 1:

BASED ON BASIC QUADRATIC EQUATION:

Find the relationship between X & Y.

1)x²-12x+32=0

y²-7x+12=0

EXPLANATION

X≤Y

PRACTICE THIS SUM :

1)X²+5X+6=0

Y²+12Y+35=0

CHECK YOUR ANSWER

The Roots of X=-2,-3

The Roots of Y=-5,-7

X>Y

2)2X²-10X+12=0

3Y²-12+9=0

CHECK YOUR ANSWER

The Roots of X=2,3;

The Roots of Y=1,3

Relationship cannot be established

#.2. TYPE 2:

BASED ON ROOTS:

This type of Pattern is asked last year in IBPS PO Mains Examination.

1)x²+9√5+100=0

y²-15√8+432=0

PRACTICE THIS SUM:

1)X²-11√8+240=0

Y²-14√8+360=0

CHECK YOUR ANSWER

The Roots of X=6√8,5√8

The Roots of Y=5√8,9√8

Cannot be Established

2)X²+11√6+90=0

Y²+15√6+216=0

CHECK YOUR ANSWER

The Roots of X=-5√6,-6√6

The Roots of Y=-6√6,-9√6

X≥Y

REMEMBER:

If you Remember this Table then you can answer without Solving,

SIGN IN THE GIVEN EQUATION SIGN OF LARGER NUMBER SIGN OF SMALLER NUMBER
+,+ – –
-,+ + +
+,- – +
-,- + –

#.3 TYPE 3:

BASED ON FINDING VALUE:

In this type you will not be given direct value’s you have to find the value from the given statement and find the relationship between them.Let see with an example,

X=Area of Square whose Side is 4cm

Y=Area of Equilateral Triangle whose side is 8cm

EXPLANATION

X=Area of Square=4²=16 Sq.cm

Y=Area of Equilateral Triangle √3/4 * 8² = 16√3 Sq.cm

Therefore X<Y.

PRACTICE THIS SUM:

1)X=Area of Triangle whose base is 1.5cm and height is 75cm

Y=Area of the equilateral triangle whose side is 12 cm

CHECK YOUR ANSWER

Area of Triangle=1/2*b*h=5625 sq.cm

Area of Equilateral Triangle=√3/4*Side²=36√3 Sq.cm

X>Y

2)X=length of the diagonal of a square whose area is 400 sq.m

Y=Perimeter of a Square Plot whose area is 400 sq.m

CHECK YOUR ANSWER

X=√(2* area)

=20√2metre

Y=√(16*400)

=4*20=80metre

X<Y

 

 

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