polinom

[stuart clark]

if  has 3  distinct real

roots and

has no real roots. then find 

[Lokman Gökçe]

Let's assume that roots of f(x) = 0 are x₁, x₂, x₃. So we have to solve that x² + 2x + 2 = x_i. (i = 1, 2, 3). Since equation x² + 2x + 2 - x_i = 0 has not real roots, then its discriminant must be negative. That is Δ = 2² - 4(2 - x_i ) < 0. Therefore x_i < 1.

for example

if f(x) = x(x + 1)(x + 2), then a = 3, b = 2, c = 0.

if f(x) = x(x + 1)(x + 3), then a = 4, b = 3, c = 0.

We can see that a, b, c aren't constant. we don't find unique value of a,b,c.