Question

Express t in terms of x: t = \frac{x^2-t+1}{x+t+1}

Answer 1

I'd rearrange and use quadratic formula

Answer 2

Attempt:
t(x+t+1) = x^2 - t + 1
xt + t^2 + 2t = x^2 + 1
t ( x + t + 2) = x^2 + 1
And then I got stuck......
Any help would be greatly appreciated!

Answer 3

I cannot use the quadratic formula, since it is not taught at that level...

Answer 4

Its a quadratic in t :
t^2 + (x+2)t -x^2 -1 = 0

Answer 5

So? I still cannot use the quadratic formula, as the ones who attempt this question have not learnt the quadratic formula yet!

Answer 6

\[t=\frac{x^2-t+1}{x+t+1} \\ \\ t(x+t+1)=x^2-t+1 \\ \\ tx+t^2+t=x^2-t+1 \\ \\ t^2+xt+2t=x^2+1 \\ \\ t^2+(x+2)t=x^2+1 \]
Completing the square method?
\[t^2+(x+2)t+\frac{(x+2)^2}{4}=x^2+1+\frac{(x+2)^2}{4} \\ \\ (t+\frac{x+2}{2})^2=x^2+1+\frac{(x+2)^2}{4} \\ \\ t=\frac{-x-2}{2}\pm \sqrt{x^2+1+\frac{(x+2)^2}{4}}\]

Answer 7

t = (x^2-t+1) / (x+t+1)
t(x+t+1) = x^2-t+1
xt + t^2 + t = x^2 - t + 1
t^2 + (x+2)t = x^2 + 1
t^2 + (x+2)t + [(x+2)/2]^2 = x^2 + 1 + [(x+2)/2]^2
{t + [(x+2)/2]}^2 = 5x^2/4 + x + 3/2
t + [(x+2)/2] =+/- sqrt(5x^2/4 + x + 3/2)
t = -x/2 - 1 +/- sqrt(5x^2/4 + x + 3/2)

That's the best I can do with completing square, but it does look ugly.

Answer 8

Yep

Answer 9

I don't suppose they know that. That's too much for them... I didn't learn this at that level as well...

Answer 10

Are you sure you wrote the problem down correctly? Does seem unusually hard for the level it's given to.

Answer 11

Like for instance what if it wasn't this:

\[t =\frac{x^2-t+1}{x+t+1}\]
but instead two parenthesis and a square was left off as an accident:

\[t =\frac{x^2-(t+1)^2}{x+t+1}\]
then it becomes:
\[t= \frac{(x+t+1)(x-t-1)}{x+t+1} = x-t-1\]
now it's super easy:
\[t=\frac{x-1}{2}\]

OK that's probably not what happened though lol.

Answer 12

Yeah, I have the same feeling, that top must be a difference of squares to make the solution simple

Answer 13

I will have to ask my math teacher to find it out tomorrow. Thanks everyone for your time!

Answer 14

Do let us know if your teacher has any interesting way to handle this. Seems unlikely though

Answer 15

Update: My teacher said it is too difficult for them as it is quadratic, therefore she cancelled the question.
Thanks everyone for your help! :)