Question

(sqrt(x)-3)^2

Answer 1

\((\sqrt{x-3^2}\))^2
by cancelling underroot with square ..
we we'll get

\(x-3^2\)=\(x-9\)

Answer 2

\(
(a-b)^2 = a^2 - 2ab + b^2 \\
(\sqrt{x}-3)^2 = (\sqrt{x})^2 - 2(\sqrt{x})(3) + 3^2 = x - 6\sqrt{x} + 9
\)

Answer 3

@ziqbal103 i think @Lemondary meant
\[\sqrt{(x-3)^{2}}\]

Answer 4

\[(\sqrt{x-3})^{2}\]

Answer 5

yes u are right may be

Answer 6

what should we type for whole square root @Mokeira
\ ( sqrt {x-3} \)

Answer 7

if the square root cover everything, the square cancels the square root so you remain with x-3

Answer 8

Well, this is what I theorized I was supposed to do:

\[(\sqrt{x} - 3 ) * (\sqrt{x} - 3 ) = \sqrt{x}^2 - 2(\sqrt{x}*3) - 3^2\]

Answer 9

i did correct @Mokeira

Answer 10

@Lemondary See the solution in my previous reply.

Answer 11

you squared 3...why @ziqbal103

Answer 12

@aum That's what I thought it was.. but it just makes my entire problem that much more complicated.
My Original problem was:

Limit of
\[(\sqrt{x} - 3)/( x - 9)\]
as x -> 9

Answer 13

I'm guessing I should multiply the bottom by this too..?

Answer 14

\(
(\sqrt{x} - 3)/( x - 9) = (\sqrt{x} - 3)/ \{(\sqrt{x} - 3) * (\sqrt{x} + 3) \} = 1 / (\sqrt{x} + 3)
\)

Answer 15

bcoz underroot will be cancellede with square root and then will be remained \(x-3^2=x-9~~{x=9}\)

Answer 16

@aum took a bit of time to see what was going on, but now I see it. Thank you.
@ziqbal103 that is not necessarily correct. It's not x-9 it's \[ x - 6\sqrt{x} + 9\]

Answer 17

\( \Large
\frac{\sqrt{x} - 3}{x - 9} = \frac{\sqrt{x} - 3}{ (\sqrt{x} - 3) * (\sqrt{x} + 3)} = \frac{1}{(\sqrt{x} + 3)} \\
\)
As x->9, the limit becomes 1 / (3+3) = 1/6

Answer 18

i was posted the comment of your 1st posted question @Lemondary

Answer 19

@ziqbal103 i think i misunderstood the initial expression...can you write it as you understood it please?

Answer 20

@ziqbal103 I know, but my first question was \[(\sqrt{x}-3)^2 \] which is \[x - 3\sqrt{x} - 3\sqrt{x} - 3^2 = x - 6\sqrt{x} + 9\]

Answer 21

is it only x that is square root? @Lemondary

Answer 22

ok @Lemondary

Answer 23

@Mokeira Yes! I think that may have been confusing in my original question. I couldn't edit it to look formatted correctly and didn't know if people didn't understand.

Answer 24

yes @aum did right question

Answer 25

oooh...now i get it...Thanks guys!

Answer 26

we w'll apply here farmula tell me how to take whole square in latex @Mokeira @aum

Answer 27

@lemondary tell me

Answer 28

yea..lets apply the formula