oops, I mean f(t)= (sq root of) (x-2) - (sq root of) (4t+1), how do I work that problem where f(t) =3?

Question

the answers are supposed to be 2,6

I'm trying it again in a different way, if I get it right before anyone else then I won't choose a best answer ahahhah, kidding

sorry, charlie, I may have written it wrong, I wish I could take a picture of it because I figured out the problem
f(x)=(sq root of) (x-2) -- (sq root of) (4t+1), and 3 is y, not x woooow

it's the long a** process of first moving either the first or the second thing over so it adds to 3

then you power up these two things, and then 9+6(sq root of)(4t+1) +4t+1=t-2 (no square root here because it was powered up)

I mean t, oops, everything is either f or t.

Answer 1

f(t) = (x-2)^½ - (4t+1)^½

one equation, two unknowns, or you have a typo.

EDIT:

f(t) = (t-2)^½ - (4t+1)^½ =3
(t-2)^½ = 3 + (4t+1)^½ rearrange
t-2 = 9 + 3(4t+1)^½ + 4t+1 square both sides
0 = 12 + 3(4t+1)^½ + 3t rearrange
0 = t + (4t+1)^½ + 4 divide by 3
-(4t+1)^½ = t + 4 rearrange
4t + 1 = t^2 + 8t +16 square both sides
0 = t^2 + 4t +15 rearrange

hopefully, I have not made an error or typo in the above, the methodology is correct.

Use Quadratic equation to solve.
http://en.wikipedia.org/wiki/Quadratic_equation

Answer 2

what's "x"?