Quadratic Equations : Step by Step Technique to Solve


Suppose we have two equations as follow:
I. Ax²+Bx+C=0
II. ay²+bx+c=0
Then from these equations we can make a master table that will help us to come to the conclusion quickly

Signs of B/b and C/c Signs of actual values of x and y
Larger value’s sign, Smaller value’s sign
+, + −, −
−, + +, +
+, − −, +
−, − +, −


Try to solve the equations instantly, if possible, by just checking whether + and – can bring us to a conclusion or not.

Example 1: 

x²+8x+12=0, +, + → −, −
y²−5y+6=0, −, + → +, +
So, using the master table, it is clear that y>x.
So, any equations which hold “+, +” and “−, +” can be quickly solved.

Example 2: 

x²+20x−32=0, +, − → −, +
y²+6y−12=0, +, − → −, +
So, from master table, it is clear that the values of x and y are both +ve and –ve. So in such cases the answer will be CND i.e. cannot be determined.


Example 1: 

x²−12x+32=0, −, + → +, +
y²−7y+12=0, −, + → +, +
In this case, we have to solve the equations. On solving, we get
x=4, x=8
y=4, y=4
Therefore x ≥ y.

Example 2: 

3x²+3x-9=0 +, - → -, +
3y²-11y+15=0 -, + → +, +

In this case we can’t reach the conclusion just by seeing only signs. So we need to solve this.


Suppose we have values -4, -.7, .3, 3.
Arrange them in a rank of decreasing order i.e. highest value will take rank 1 then after next value will get rank 2 and so on…
4= RANK1, .3=RANK2, -.7=RANK3, -4=RANK4.
When the ranks of values of x and y is:
x=1, 2
y=3, 4
Then obviously, x > y.
When the ranks of values of x and y is:
y=1, 2
x=3, 4
Then obviously, y > x.
Any other ranks combinations will give CND.
But when the ranks get tied up, then
x= -7, -3
y= -3, -2
So X’s rank would be 4, 2
Y’s rank would be 2, 1. (We have assumed rank 2 and rank 3 equal to rank 2.)
Then we compare them and we get;
x=y and y>x. On combining both these results we reach to a desired result i.e. y ≥ x.
NOTE: If we ever have to combine x>y and x<y then the result will obviously be CND.

Some special cases:


Answer in this case will be CND as a square would always gives a +ve as well as –ve values.
x= 30, -30
y= 30, -30
So any two squares and their variations will always be CND, irrespective of the values.


Result will be CND. Because, we ultimately will get +ve as well as –ve values.
x= ± 625)(1⁄4)= ±5
y2= 25
y= ±5
Option will be CND.
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