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How to find quickly square root of a perfect square polynomial of degree 6?

How to find quickly square root of a perfect square polynomial of degree 6? This is a little method which will enable you to get the answer quickly. Let the polynomail be Px^6+Qx^5+Rx^4+Sx^3+Tx^2+Ux+V. Let its root be ax^3+bx^2+cx+d. a= square root of P. b= Q/2a c= (R-b^2)/2a d= U/2c

Coordinate geometry trick: If co-ordinates of two points on a line are given and you are asked to find the equation of the line. How to find it quickly.

 Coordinate geometry trick:

If co-ordinates of two points on a line are given and you are asked to find the equation of the line. How to find it quickly.

0) If Y2/X2 = Y1/X1 = k

the required equation is y = kx.

or kx-y=0.

Example 1:

A (2,4)

B (1,2)

Here it is quite easy to understand that k = 4/2 = 2/1 = 2.

so y = 2x

or 2x-y=0 is the required equation of the line.

*Or*

1) Write first the coordinates of that point whose y- co-ordinate is larger than y-co-ordinate of the other point.

2) Below it, write the co-ordinates of the other point.

3) Find the difference between the Y- co-ordinates. You have got now the coefficient of the x in the required equation of the line.

4) find the negative of the difference between the X coordinates in the same order. This is the coefficient of y in the required equation of the line.

5) Now X2 , Y2

              X1 , Y1.

Find + ( X2 Y1 - X1 Y2 ).

It is the constant part of the equation.

5a) write the answers obtained steps 3), 4) and 5) with x, y and appropriate signs for the constant, and equate it to zero.

6) This is the required equation of the line in a x + b y + c = 0 form. *Refer Example 2 given below.*

7) If in a x + b y + c = 0, a, b & c not coprimes, then divide the equation by the highest common factor (HCF) of a,b,c.

This is your required equation. *Refer Example 3 given below.*

Example 2:

A(3,-2)

B(4,-4)

-2>-4 

A(3,-2)

B(4,-4)

[-2-(-4)] x - (3-4) y + (3x-4 - 4x-2) =0

2x+y+(-12+8)=0

2x+y=4.

is the required equation.

Example 3:

B (1,-1)

A (7,7)

7 > -1

A (7,7)

B (1,-1)

[(7 - (-1) ] x -(7 - 1)y + (7 x - 1 - 7 x 1) = 0

8X - 6y -14 =0

8, 6 and 14 are not coprimes and their HCF is 2 therefore dividing the equation by 2 we get,

4x -3y - 7 = 0

or 4x -3y = 7 is the required equation of the line passing through the given points.

Hope you liked the post. It is more useful for 11 Std Science student who are preparing for CET or JEE.