Question

Sa se determine toate valorile nenule ale parametrului real a astfel încât ecuația sa aibă cel puțin o rădăcina reala.


Mulțumesc :)

Answer 1

Răspuns:

$\frac{1\pm \sqrt{2}}{2}$

Explicație pas cu pas:

$\sqrt{x-2} + \sqrt{ax^2 - 2x - \frac{1}{a}} = 0\\ \\ \sqrt{v} \geq 0, \forall v \in \mathbb{R}_+ \\ \\ \implies \sqrt{x-2} = \sqrt{ax^2 - 2x - \frac{1}{a}} = 0\\ \\ \sqrt{x-2} = 0 \implies x-2 = 0 \implies x = 2\\ \\ \sqrt{ax^2 - 2x - \frac{1}{a}} = 0\\ \\ \sqrt{a\cdot 2^2 - 2\cdot 2 - \frac{1}{a}} = 0\\ \\ 4a - 4 - \frac{1}{a} = 0\Bigg |\cdot a\\ \\ 4a^2 - 4a - 1 = 0\\ \\ \Delta = 16 - 4\cdot 4 \cdot (-1) = 16 + 16 = 32\\ \\ \sqrt{\Delta} = \sqrt{32} = 4\sqrt{2}\\ \\ a_{1,2} = \frac{4 \pm 4\sqrt{2}}{8} = \boxed{\frac{1\pm \sqrt{2}}{2}}$

Answer 2

$\sqrt{x-2}+\sqrt{ax^2-2x-\dfrac{1}{a}} = 0\\ \\\\ \sqrt{x-2}\geq 0 \\\\ \sqrt{ax^2-2x-\dfrac{1}{a}} \geq 0\\ \\ \\ \Rightarrow \sqrt{x-2} = 0\quad \text{si}\quad \sqrt{ax^2-2x-\dfrac{1}{a}}=0\\\\\\\Rightarrow x-2 = 0\Rightarrow x = 2\\ \\ \Rightarrow \sqrt{a\cdot 2^2-2\cdot 2-\dfrac{1}{a}} = 0 \Rightarrow \sqrt{4a-4-\dfrac{1}{a}} = 0\Rightarrow \\ \\ \Rightarrow 4a-\dfrac{1}{a}-4 = 0 \Rightarrow 4a^2-4a-1 = 0$

$\Delta = 16+16 = 32 \Rightarrow a_{1,2} =\dfrac{4\pm 4\sqrt 2}{8} = \dfrac{1\pm \sqrt 2}{2}\\ \\\\ \Rightarrow a \in \Bigg\{\dfrac{1-\sqrt 2}{2},\, \dfrac{1+\sqrt 2}{2}\Bigg\}$