Nvm I got that one by actually trying. This one I'm stuck on though:
2(x^2x)+5sqrt(x^2+2x)+2=0
Name:
Anonymous2007-09-18 12:37 ID:aIg3VJeP
whats the sqrt stand for if you can tell me that i can probally help you.
Name:
Anonymous2007-09-18 12:38 ID:NqgyPcsD
Sqrt stands for square root
It's basically 5 with the square root of x^2+2x next to it.
It would be awesome if you showed me how you did it as well. I have a quiz on that tommorow.
Name:
Anonymous2007-09-18 12:39 ID:aIg3VJeP
ok let me get a pen and papper and ill be at it in a sec.
Name:
Anonymous2007-09-18 12:43 ID:aIg3VJeP
is the 5 and the square root being multiplied?
Name:
Anonymous2007-09-18 12:57 ID:aIg3VJeP
ok so far i got to this point starting from original equation
2(x^2x)+5sqrt(x^2+2x)+2=0
ok then i did this
[2(x^2x)+5sqrt(x^2+2x)+2=0]^2
this eliminates the sqrt leaving this
2^2(x^2x)^2+5^2(x^2+2x)+2^2=0
then simplify
4(x^4x)+25(x^2+2x)+4=0
see where this is going?
Name:
Anonymous2007-09-18 13:03 ID:NqgyPcsD
I made a small mistake in the equation
2(x^2+2x)+5sqrt(x^2+2x)+2=0
In the first parantheses
Name:
Anonymous2007-09-18 13:04 ID:aIg3VJeP
oh that makes this easier one sec
Name:
Anonymous2007-09-18 13:21 ID:aIg3VJeP
ok i only got so far but then i forgot what to do next but ill show you what i have.
2(x^2+2x)+5sqrt(x^2+2x)+2=0
now
2(x^2+2x)+5[sqrt(x^2+2x)^2]+2=0
that leaves you with
2(x^2+2x)+5(x^2+2x)+2=0
simplify
(2x^2+2x)+(5x^2+2x)+2=0
then
(7x^2+2x)+2=0
can you solve from there?
Name:
Anonymous2007-09-18 13:23 ID:NqgyPcsD
Thank so much
Name:
Anonymous2007-09-18 13:24 ID:aIg3VJeP
no problem and rember to check the math by plugging in the answer into the original equation.
Name:
Anonymous2007-09-18 14:07 ID:ZRWrqfvr
you guys are such stereotypical redneck hicks man, stop embarrassing yourselves