I keep getting this wrong (1 Viewer)

[Mumma]

3t / (t^2 - 6) = root(3)

solve... (find t)

It seems so easy but I keep getting it wrong, over and over again. I seriously have no idea why. I just want to make sure its not a printing error at the end of the book or something.

 

[Templar]

It seems so easy but I keep getting it wrong, over and over again. I seriously have no idea why. I just want to make sure its not a printing error at the end of the book or something.

Well what did you get? And what does the book say?

 

[Templar]

-sqrt3 and 2sqrt3

 

[Mumma]

Can you post the steps. Im going crazy. Im doing something wrong and I dont know what it is.

 

[nick1048]

I get 42/root3 and -39/root3... hmmm

 

[Slidey]

3t / (t^2 - 6) = root(3)

sqrt3(t^2-6)=3t
[sqrt3(t^2-6)]^2=(3t)^2
t^4-15t^2+36=0
t^2={15+sqrt(225-144)}/2
t^2=(15+9)/2=12 or 3
t=+2sqrt3 or +sqrt3
Then you just substitue back into the original equation and discard non-solutions. I just couldn't be bothered solving the original thing while it contained a surd.

EDIT: Discarding the solutions:
t=+2sqrt3:
pos:
3t / (t^2 - 6) = 3.2sqrt3/ (12 - 6) = sqrt3 WORKS
neg obiously doesn't work, by inspection.
t=+sqrt3:
pos:
3t / (t^2 - 6) =3.sqrt3/(3-6)=-sqrt3 Doesn't work
neg: again by inspection, neg will work.

so you have: t = 2sqrt3 and -sqrt3

 

[Templar]

I just couldn't be bothered solving the original thing while it contained a surd.

Neither could I. Cheated and just plugged it in Mathematica.

 

[nick1048]

Im not even ganna pretend to understand that... -_-'

 

[QuaCk]

just divide both sides by root(3) and solve quadratically.
and keep in mind the denominator can't = 0

 

[Slidey]

Im not even ganna pretend to understand that... -_-'

It's just the freaking quadratic formula! You can even solve the quartic produced by inspection for god's sake! (3+12=15 and 3.12=36)

 

[QuaCk]

3t / (t^2 - 6) = root(3)
root(3).t / (t^2 - 6) = 1
root(3).t = t^2 - 6 ( t^2 - 6 cannot be 0 )
t^2 - root(3).t - 6 = 0

then use quadratic formula root thingy

 

[Mumma]

Can someone tell me why I cant do this? -_-

3t / (t² - 6) = √3

3t = (√3)t² - 6√3

and then like do the quadratic formula on this pos?

(√3)t² - 3t - 6√3 = 0

isnt that

ax² - bx - c ... ???

t = [ 3 +- √(9 - 4*√3*-6√3) ] / 2√3

t = (3 +- 9) / 2√3

I mean what am I doing illegal here? Excuse ... just new year 11 student...

 

[Templar]

t = (3 +- 9) / 2√3

I mean what am I doing illegal here? Excuse ... just new year 11 student...

Nothing. Just rationalise the denominator.

 

[QuaCk]

rationalise?

dammit, too slow -.-

 

[Mumma]

HAHAHAHAHAAHAH

oh my god
I spent an hour doing it over and over and over again, always getting the same answer.
going crazy. turned out i was right all along. fucking hilarious.

goddamnit there go 2 hours of my life.

Oh yeah, and thanks a TON guys. Really helpful.