Solve the equation (\frac{t}{t+1})^{2}-\frac{2t}{t+1}-8=0

arenceabigns 2021-09-23 Answered
Solve the equation
(tt+1)22tt+18=0
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Expert Answer

SkladanH
Answered 2021-09-24 Author has 80 answers
Step 1
The given equation is,
(tt+1)22tt+18=0
On simplifying the given equation, we get
(tt+1)22tt+18=0
(tt+1)22(tt+1)8=0
Let, tt+1=u,t1
Then the given equation becomes,
u22u8=0
Step 2
On solving the obtained quadratic equation, we get
u22u8=0
u24u+2u8=0
u(u-4)+2(u-4)=0
(u+2)(u-4)=0
u=-2 or 4
Step 3
When u=−2
tt+1=2
t=-2(t+1)
t=-2t-2
t+2t=-2
3t=-2
t=23
When u=4
tt+1=4
t=4(t+1)
t=4t+4
t-4t=4
-3t=4
t=43
Therefore, the values of t are 23 and 43
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