Question

Pentru ce valori ale parametrului real m, m ≠ 6 ecuaţia (m + 6)x2

– 4mx + m + 1 = 0 are

a. rădăcini opuse;
b. o rădăcină nulă;
c. rădăcini inverse;
d. o rădăcină -1;
e. rădăcini egale.

Answer 1

$m \in \mathbb{R}\\(m-6)x^{2} -4mx + (m+1) = 0$

$\text{Fie x_{1}, x_{2} - solutiile ecuatiei}$

Conform relațiilor lui Viete:  

$x_{1} + x_{2} = \frac{4m}{m+6}\\x_{1}x_{2} = \frac{m+1}{m+6}$

a) $x_{1} = -x_{2} \implies x_{1} + x_{2} = 0 \iff \frac{4m}{m+6} = 0 \implies \boxed{m = 0}$

b) $x_{1} = 0 \implies x_{1}x_{2} = 0 = \frac{m+1}{m+6} \implies \boxed{m = -1}$

c) $x_{1} = -1 \implies x_{1}x_{2} = -x_{2} = \frac{m+1}{m+6} \implies x_{2} = -\frac{m+1}{m+6}\\x_{1} + x_{2} = x_{2} - 1 = \frac{4m}{m+6} = \frac{-m - 1}{m + 6} - 1\\\frac{-m - 1 -m - 6}{m + 6} = \frac{-2m - 7}{m + 6} = \frac{4m}{m + 6}\\-2m - 7 = 4m \implies \boxed{m = -\frac{7}{6}}$

d) $x_{1} = x_{2} \implies x_{1}x_{2} = x_{1}^2 = \frac{m + 1}{m + 6}.\\x_{1} + x_{2} = 2x_{1} = \frac{4m}{m + 6} \iff x_{1} = \frac{2m}{m + 6} \iff x_{1}^{2} = \frac{4m^2}{(m+6)^2}\\\text{Din egalitatea anteriora: } x_{1}^2 = \frac{4m^2}{(m+6)^2} = \frac{m+1}{m+6}\\\frac{4m^{2}}{(m+6)^2} = \frac{(m+1)(m+6)}{(m+6)^2} = \frac{m^2 + 7m + 6}{(m+6)^2}\\4m^{2} = m^2 + 7m + 6$

$3m^{2} - 7m - 6 = 0\\\boxed{m_{1, 2} = \frac{7 \pm \sqrt{49 - 4(3)(-6)}}{6} = \frac{7 \pm \sqrt{11^{2}}}{6} = \frac{7 \pm 11}{6} \in \{3, -\frac{2}{3}\}}$