Given that 1/x1+1/x2=2, what is t such as x^2-(t-2)x+t+1=0?

It is given that 1/x1 + 1/x2 =
2.

t has to be determined such that x^2 - (t-2)x + t +1 =
0

1/x1 + 1/x2 = 2

=>
(x1 + x2)/x1*x2 = 2

For a quadratic equation ax^2 + bx + c
= 0 with roots x1 and x2,

c/a = x1*x2 and -b/a = x1 +
x2

Here the equation is x^2 - (t-2)x + t +1 =
0

=> (t - 2)/(t + 1) =
2

=> t - 2 = 2t +
2

=> t =
-4

The required value of t =
-4