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.