SOLUTION: (t/t+1)^2-(t/t+1)-12=0

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Question 949097: (t/t+1)^2-(t/t+1)-12=0
Found 2 solutions by josgarithmetic, MathTherapy:
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
The meaning of what you wrote is %28t%2Ft%2B1%29%5E2-%28t%2Ft%2B1%29-12=0.


Certainly you wanted (t/(t+1))^2-(t/(t+1))-12=0
which will appear rendered as %28t%2F%28t%2B1%29%29%5E2-%28t%2F%28t%2B1%29%29-12=0.

Multiply the left (and right) sides by %28t%2B1%29%5E2 which is the common denominator. You will have a quadratic equation, which you will simplify into general form or possibly be able to factorize.

Answer by MathTherapy(10551) About Me  (Show Source):
You can put this solution on YOUR website!

(t/t+1)^2-(t/t+1)-12=0
%28t%2F%28t+%2B+1%29%29%5E2+-+%28t%2F%28t+%2B+1%29%29+-+12+=+0

Let a = %28t%2F%28t+%2B+1%29%29

Then %28t%2F%28t+%2B+1%29%29%5E2+-+%28t%2F%28t+%2B+1%29%29+-+12+=+0 becomes: a%5E2+-+a+-+12+=+0

(a - 4)(a + 3) = 0

a - 4 = 0           OR           a + 3 = 0

a = 4               OR           a = - 3



4+=+%28t%2F%28t+%2B+1%29%29 ------- Substituting 4 for a

4(t + 1) = t ------ Cross-multiplying

4t + 4 = t

4t - t = - 4

3t = - 4

highlight_green%28t+=+-+4%2F3%29



-+3+=+%28t%2F%28t+%2B+1%29%29 ------- Substituting - 3 for a

- 3(t + 1) = t ----- Cross-multiplying

- 3t - 3 = t

- 3t - t = 3

- 4t = 3

highlight_green%28t+=+-+3%2F4%29