### STEP-1

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 |
---|---|

+, + | −, − |

−, + | +, + |

+, − | −, + |

−, − | +, − |

### STEP-2

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.

### STEP-3

#### 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.

### STEP-4

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

Suppose,

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:

#### 1.

x²=900y²=900

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.

#### 2.

x2-240=460y2+500=1000

Result will be CND. Because, we ultimately will get +ve as well as –ve values.

NOTE:

x4=625

x= ± 625)(1⁄4)= ±5

y2= 25

y= ±5

Option will be CND.

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