What are the coefficients of the function tx^2+(t-1)x-t+2 ?

Since it is not specified if the roots of the equation are
real, distinct, equal or imaginary, we'll choose to solve the problem, considering that
f(x) has 2 distinct real roots.

The equation has 2 distinc
real roots if and only if the discriminant delta is strictly
positive.

delta = b^2 -
4ac

We'll identify a, b and
c.

a = t

b = t -
1

c = 2 - t

delta = (t-1)^2 -
4*t*(2 - t)

We'll impose the constraint: delta >
0

(t-1)^2 - 4*t*(2 - t) >
0

We'll expand the square and remove the
brackets:

t^2 - 2t + 1 - 8t + 4t^2 >
0

We'll combine like
terms:

5t^2 - 10t +
1>0

Now, we'll determine the roots of the expression
5t^2 - 10t + 1.

5t^2 - 10t + 1 =
0

We'll apply the quadratic
formula:

t1 =
[10+sqrt(100-20)]/10

t1 =
(10+4sqrt5)/10

t1 =
(5+2sqrt5)/5

t2 =
(5-2sqrt5)/5

The expression is positive outside the roots,
namely when t belongs to the intervals:(-infinite , (5-2sqrt5)/5) U ((5+2sqrt5)/5 ,
+infinite).

The coefficients of the equation could be
calculated choosing values for t from intervals :(-infinite , (5-2sqrt5)/5) U
((5+2sqrt5)/5 , +infinite).

For  t = -1, the
coefficients of the equation are: a = -1; b = -1 - 1 = -2; c = 2 + 1 =
3.