Question

Hey just wondering if I can factor out the x in:

sqrt(9x^2+x) +3x

Answer 1

\[\sqrt{9x^{2} + x } + 3x\]

Answer 2

What you could do is \[\sqrt{9x^2+x}=\sqrt{x(9x+1)}=\sqrt{x}\sqrt{9x+1}\]

Answer 3

hm okay thanks, i don't think that'd help me with this problem ty

Answer 4

lol no problem :)

Answer 5

wait i can't factor the 3 right?

Answer 6

Nope, what are you trying to do though?

Answer 7

Well i'm trying to think of ways to solve my limit problem

Answer 8

you mind taking a look at it and giving ur advice?

Answer 9

sure

Answer 10

\[\lim \sqrt{9x^{2}+ x} -3x\]
as x --> infinity

Answer 11

I start by multiplying by the conjugate

Answer 12

which led me to trying to find out if i can factor the denominator or not.

Answer 13

just finding it so difficult to understand the answer which is \[\frac{ 1 }{ 6 }\]

Answer 14

Well you're right to start of by multiplying by the conjugate. After doing so, however, you can use the properties of a limit to rewrite it as the quotient of the limits rather than the limit of the quotient.

Answer 15

Oh wow I think I got it?
After staring at this problem for like 45 minutes I have:
\[\frac{ 9x-8 }{ \sqrt{9} +3 }\]

Answer 16

that extra x is killing me

Answer 17

How did you get to that expression?

Answer 18

After multiplying by the conjugate i get:
\[\frac{ 9x^{2} - 8x }{ \sqrt{9x^{2}+x} + 3x }\]

Answer 19

then i multiply the top by the highest power in the denominator \[\frac{ 1 }{ x }\] for the numerator and: \[\sqrt{\frac{ 1 }{ x^{2} }}\]

Answer 20

which leads me to the expression i got

Answer 21

when you multiply the conjugate your numerator should simplify to x since (sqrt(9x^2+x)-3x) * (sqrt(9x^2+x) + 3x) = x so you will have \[\frac{ x }{ \sqrt{9x^2+x}+3x }\] and from there you can multiply by 1/x.

Answer 22

OK thanks again twis7ed