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

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### 1 Answer

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.**