arenceabigns
2021-09-23
Answered

Solve the equation

${\left(\frac{t}{t+1}\right)}^{2}-\frac{2t}{t+1}-8=0$

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SkladanH

Answered 2021-09-24
Author has **80** answers

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$-\frac{2}{3}\text{}{\textstyle \phantom{\rule{1em}{0ex}}}\text{and}{\textstyle \phantom{\rule{1em}{0ex}}}\text{}-\frac{4}{3}$

The given equation is,

On simplifying the given equation, we get

Let,

Then the given equation becomes,

Step 2

On solving the obtained quadratic equation, we get

u(u-4)+2(u-4)=0

(u+2)(u-4)=0

u=-2 or 4

Step 3

When u=−2

t=-2(t+1)

t=-2t-2

t+2t=-2

3t=-2

When u=4

t=4(t+1)

t=4t+4

t-4t=4

-3t=4

Therefore, the values of t are

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