Question

I just forgot the way to find the rationalizing factor can any1 help me with this?

Answer 1

I just want an example for this

@Ganpat @jagatuba and @saifoo.khan

Answer 2

okay i will

Answer 3

Hmn please go ahead @Nick_Black

Answer 4

http://www.twiddla.com/895293
come here

Answer 5

You there?.

Answer 6

@saifoo.khan can u help/?

Answer 7

Isn't @Nick_Black helping?

Answer 8

aap hain yaha pe?? @saifoo.khan

Answer 9

But i didn't get how to get rationalizing factor @Nick_Black ..

Answer 10

u rationalize the denominator this is to change it frm sqrt to a whole no.

Answer 11

@mathslover :

Rationalisation is multiply and divide by denominator... so as to remove any square root sign in the denominator... This is just to simplify further calculations..

Answer 12

And the denominator is denoted as factor in the equation.. SO rationalizing factor..

Answer 13

ganpat is righjt here

Answer 14

Oh k!!! thanks a lot

Answer 15

a summary:
\[\frac{a}{\sqrt b} \implies \frac{a}{\sqrt b} \times \frac{\sqrt b}{\sqrt b} \implies \frac{a\sqrt b}{b}\]

Answer 16

Nice pic @lgbasallote

Answer 17

is this nice photo dude

Answer 18

This is the case when denominator is a single term:

If it is of two terms then:

We always change the sign of one out of them which you can think can make my denominator simple.. For example:

Suppose you have:

\[\frac{a}{x - \sqrt{y}}\]

Change the sign of \(\sqrt{y}\) so that you can use : \((m-n)(m+n) = m^2 - n^2\)

So you will multiply and divide by \(x + \sqrt{y}\) ( '-' converted to '+' )

\[\frac{a}{{x - \sqrt{y}}} \implies \frac{a}{{x - \sqrt{y}}} \times \frac{{x + \sqrt{y}}}{{x + \sqrt{y}}} \implies \frac{ax + a \sqrt{y}}{{x^2 - y}}\]


Now here one more thing I want to tell you: You noticed that by using \({x + \sqrt{y}}\) as Rationalization Factor you will get \(x^2 - y\) : The y term is negative..


But in case of complex number when you rationalize in the same way, then you will get: \(x^2 + y\)..

I show you:

Take the same example but suppose you have complex number in denominator: \({x - i \cdot\sqrt{y}}\)..

So:

\[\frac{a}{{x - i \cdot \sqrt{y}}} \implies \frac{a}{{x - i \cdot\sqrt{y}}} \times \frac{{x + i \cdot \sqrt{y}}}{{x + i \cdot \sqrt{y}}} \implies \frac{ax + a \cdot i \cdot \sqrt{y}}{{x^2 - i^2 \cdot (\sqrt{y}})^2} \implies \frac{ax + a \cdot i \cdot \sqrt{y}}{{x^2 + y}}\]

Since \(i^2 = -1 \) which makes it positive...

Answer 19

isnt that called conjugation? is it still under rationalization? o.O

Answer 20

Yes...

Answer 21

hmm yup it is

Answer 22

Rationalization basically means that to rationalize the denominator..

Don't you think we here making our denominator rational...??

Answer 23

means making the denominator a whole number, isn't it??
@waterineyes

Answer 24

Yes very right...

Or it can be rational too..

Answer 25

such as