Question

Exercitiul 3, va rog!!! Ofer 100 de puncte!

Answer 1

Explicație pas cu pas:

$f(x) = (a + 4) x^{2} + (2a + 5) x + a - 4$

a)

$a = 2, \ f(x) = 0$

$f(x) = (2 + 4) x^{2} + (4 + 5) x + 2 - 4 = 6 {x}^{2} + 9x - 2$

$6 {x}^{2} + 9x - 2 = 0$

$\Delta = 81 + 48 = 129 \\ x_{1} = \frac{-9 - \sqrt{129} }{12} \\ x_{2} = \frac{-9 + \sqrt{129} }{12}$

b)

$a = 2, \ f(x) \leqslant 0$

$6 {x}^{2} + 9x - 2 \leqslant 0$

$\implies x \in \Big[\frac{-9 - \sqrt{129} }{12} ;\frac{-9 + \sqrt{129} }{12}\Big]$

c)

$x^{2} - Sx + P = 0$

$(a + 4) x^{2} + (2a + 5) x + a - 4 = 0$

$x^{2} - \Big(- \frac{2a + 5}{a + 4}\Big) \cdot x + \frac{a - 4}{a + 4} = 0$

$S = x_{1} + x_{2} \implies S = - \frac{2a + 5}{a + 4}$

$P = x_{1} \cdot x_{2} \implies P = \frac{a - 4}{a + 4}$

$x_{1} + x_{2} = -4 \cdot x_{1} \cdot x_{2} \implies S = -4 \cdot P$

$- \frac{2a + 5}{a + 4} = - 4 \cdot \frac{a - 4}{a + 4} \\ 2a + 5 = 4(a - 4) \\ 2a + 5 = 4a - 16 \\ 4a - 2a = 5 + 16 \\ 2a = 21 \implies a = \frac{21}{2} \\ \iff a = 10.5$