Question

Sa se determine m astfel incat radacinile ecuatiei x^2(m-1)-(2m-1)x-(m+1)=0 sa existe relatia: x1^2+x2^2=17.​

Answer 1

$(m-1)x^2 - (2m-1)x - (m+1) = 0 \\x_{1}^2 + x_{2}^2 = (x_{1} + x_{2})^2 - 2x_1 x_{2} = 17\\\text{Conform relatiilor Viete: } x_{1} + x_{2} = \frac{2m - 1}{m - 1} \text{, } x_1 x_2 = \frac{-(m+1)}{m-1}.$

De aici avem:

$x_1^2 + x_2^2 = (\frac{2m - 1}{m - 1})^2 - 2(\frac{-(m+1)}{m - 1}) = \frac{4m^2 - 4m + 1}{(m -1)^2} + \frac{2m + 2}{m - 1} = \frac{4m^2 - 4m + 1 + (2m + 2)(m-1)}{(m-1)^2}$

$\frac{4m^2 - 4m + 1 + 2m^2 -2m + 2m - 2}{(m-1)^2} = \frac{6m^2 - 4m - 1}{(m-1)^2} = 17\\6m^2 - 4m - 1 = 17m^2 - 34m + 17\\11m^2 - 30m + 18 = 0\\\\m_{1, 2} = \frac{30 \pm \sqrt{30^2 - 4(11)(18)}}{22} = \frac{30 \pm \sqrt{2^2 \cdot 3^3}}{22} = \frac{30 \pm 6\sqrt{3}}{22} = \frac{15 \pm 3\sqrt{3}}{11}$

Răspuns: $\boxed{m \in \{\frac{15 \pm 3\sqrt{3}}{11}\}}$