Thread: expanding brackets

Hi what is (3x + 1/x)^2 ???????

Originally Posted by wolfhound

Hi what is (3x + 1/x)^2 ???????

HINT: (a + b)^2 = (a + b) times (a + b)

Originally Posted by Wilmer

HINT: (a + b)^2 = (a + b) times (a + b)

I know that

Hi!

In addition to Wilmer's post

One should lnow the binimial theorems

$\displaystyle (a+b)^2 = a^2 + 2ab + b^2$

$\displaystyle (a-b)^2 = a^2 - 2ab + b^2$

$\displaystyle (a+b)(a-b) = a^2-b^2$

In the example it is a = 3x and b = 1/x

Can you solve it now?

Edit

Originally Posted by wolfhound

I know that

Okay, here is the solution

(3x)^2+2*(3x)*(1/x)+(1/x)^2

$\displaystyle = 9x^2+\frac{6x}{x} + \frac{1^2}{x^2}$

$\displaystyle = 9x^2 +6 + \frac{1}{x^2}$

I know all that pascals triangle stuff ,what I dont understand is what becomes of 3x when its multiplied by 1/x?? or 1/x x 1/x?

Okay, here is the solution

(3x)^2+2*(3x)*(1/x)+(1/x)^2

$\displaystyle = 9x^2+\frac{6x}{x} + \frac{1^2}{x^2}$

$\displaystyle = 9x^2 +6 + \frac{1}{x^2}$[/QUOTE]

thanks

Originally Posted by wolfhound

I know all that pascals triangle stuff ,what I dont understand is what becomes of 3x when its multiplied by 1/x?? or 1/x x 1/x?

$\displaystyle (3x)(\frac{1}{x})= \frac{3x}{x}= 3$

$\displaystyle (\frac{1}{x})(\frac{1}{x})= \frac{1}{x^2}$