Question

Solve the equation 1/t-2=t/8

Answer 1

Two solutions were found :

t =(16-√288)/-2=8+6√ 2 = 0.485

t =(16+√288)/-2=8-6√ 2 = -16.48

Explanation:

Answer 2

-IF THE EQUATION IS
`(1)/(t-2)=(t)/(8)`, THEN:

`t_1=4\\t_2=-2`

-IF THE EQUATION IS
`(1)/(t)-2=(t)/(8)`, THEN:

`t_1=-16.485\\t_2=0.485`

Explanation:

-IF THE EQUATION IS
`(1)/(t-2)=(t)/(8)` THE PROCEDURE IS:

Multiply both sides of the equation by
`t-2` and by 8:

`(8)(t-2)((1)/(t-2))=((t)/(8))(8)(t-2)\\\\(8)(1)=(t)(t-2)\\\\8=t^2-2t`

Subtract 8 from both sides of the equation:

`8-8=t^2-2t-8\\\\0=t^2-2t-8`

Factor the equation. Find two numbers whose sum be -2 and whose product be -8. These are -4 and 2:

`0=(t-4)(t+2)`

Then:

`t_1=4\\t_2=-2`

-IF THE EQUATION IS
`(1)/(t)-2=(t)/(8)` THE PROCEDURE IS:

Subtract
`(1)/(t)` and
`2`:

`(1)/(t)-2=(t)/(8)\\\\(1-2t)/(t)=(t)/(8)`

Multiply both sides of the equation by
`t`:

`(t)((1-2t)/(t))=((t)/(8))(t)\\\\1-2t=(t^2)/(8)`

Multiply both sides of the equation by 8:

`(8)(1-2t)=((t^2)/(8))(8)\\\\8-16t=t^2`

Move the
`16t` and 8 to the other side of the equation and apply the Quadratic formula. Then:

`t^2+16t-8=0`

`t=(-b\±√(b^2-4ac))/(2a)\\\\t=(-16\±√(16^2-4(1)(-8)))/(2(1))\\\\t_1=-16.485\\t_2=0.485`