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Piyasa Kitaplarındaki Hatalı Sorular => Hatalı Cebir Soruları => Konuyu başlatan: stuart clark - Haziran 18, 2011, 05:56:57 ös

Başlık: polinom
Gönderen: stuart clark - Haziran 18, 2011, 05:56:57 ös

if  (http://latex.codecogs.com/gif.latex?f(x)%20=%20x^3+ax^2+bx+c$) has 3  distinct real

roots and (http://latex.codecogs.com/gif.latex?\left(x^2%20+2x+2\right)^3+%20a\left(x^2+2x+2\right)^2%20+b\left(x^2+2x+2\right)+c$)

has no real roots. then find  (http://latex.codecogs.com/gif.latex?a,b,c$)

Başlık: Ynt: polinom
Gönderen: Lokman Gökçe - Haziran 19, 2011, 03:04:15 ös

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.