Răspuns :

Răspuns:

Explicație pas cu pas:

Intersectia cu Ox are y = 0

y = f(x)

(a + 1)x^2 + 3(a - 1)x + a - 1 = 0

ecuatia are doua radacini reale daca Δ > 0

Δ = 9(a - 1)^2 - 4(a + 1)(a - 1)

= 9(a^2 - 2a + 1) - 4(a^2 - 1)

= 9a^2 - 18a + 9 - 4a^2 + 4

= 5a^2 - 18a + 13

5a^2 - 18a + 13 > 0

Δ = 524 - 260 = 264

a1 = (18 + √264)/10

a2 = (18 - √264)/10

a ∈ (-∞, a2) U (a1, + ∞)

Targoviste44id

[tex]\it Gf\cap Ox=\{x_1,\ x_2\},\ unde\ x_1,\ x_2\ sunt\ solu\c{\it t}ii\ pentru\ ecua\c{\it t}ia\ f(x)=0.\\ \\ Cele\ dou\breve{a}\ solu\c{\it t}ii\ sunt\ reale\ distincte\ dac\breve{a}\ \Delta >0.\\ \\ \Delta >0\ \Leftrightarrow [3(a-1)]^2-4(a+1)(a-1) >0 \Leftrightarrow 9(a-1)^2-4(a+1)(a-1)>0\Leftrightarrow\\ \\ \Leftrightarrow(a-1)(9a-9-4a-4) >0\Leftrightarrow (a-1)(5a-13) >0 \Leftrightarrow \\ \\ \Leftrightarrow a\in(-\infty;\ \ 1)\cup(2,6;\ \infty)[/tex]

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