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

Solve the equation
${\left(\frac{t}{t+1}\right)}^{2}-\frac{2t}{t+1}-8=0$
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Step 1
The given equation is,
${\left(\frac{t}{t+1}\right)}^{2}-\frac{2t}{t+1}-8=0$
On simplifying the given equation, we get
${\left(\frac{t}{t+1}\right)}^{2}-\frac{2t}{t+1}-8=0$
${\left(\frac{t}{t+1}\right)}^{2}-2\left(\frac{t}{t+1}\right)-8=0$
Let, $\frac{t}{t+1}=u,t\ne -1$
Then the given equation becomes,
${u}^{2}-2u-8=0$
Step 2
On solving the obtained quadratic equation, we get
${u}^{2}-2u-8=0$
${u}^{2}-4u+2u-8=0$
u(u-4)+2(u-4)=0
(u+2)(u-4)=0
u=-2 or 4
Step 3
When u=−2
$\frac{t}{t+1}=-2$
t=-2(t+1)
t=-2t-2
t+2t=-2
3t=-2
$t=-\frac{2}{3}$
When u=4
$\frac{t}{t+1}=4$
t=4(t+1)
t=4t+4
t-4t=4
-3t=4
$t=-\frac{4}{3}$
Therefore, the values of t are