Explicație pas cu pas:
[tex]x^{2} - 2x + m + 1 = 0[/tex]
[tex]a = 1; \: b = -2 ; \: c = m + 1[/tex]
[tex]Δ \geqslant 0 = > ( - 2)^{2} - 4(m + 1) \geqslant 0 \\ 4 - 4m - 4 \geqslant 0 \\ 4m \leqslant 0 = > m \leqslant 0[/tex]
[tex]x_{1} + x_{2} = \frac{ - b}{a} = \frac{ - ( - 2)}{1} = 2[/tex]
[tex]x_{1}x_{2} = \frac{c}{a} = \frac{m + 1}{1} = m + 1[/tex]
a)
[tex]x_{1} = 3x_{2}[/tex]
[tex]3x_{2} + x_{2} = 2 \\ 4x_{2} = 2 = > x_{2} = \frac{1}{2} [/tex]
[tex]x_{1} = 3 \times \frac{1}{2} = \frac{3}{2} [/tex]
[tex]x_{1}x_{2} = \frac{1}{2} \times \frac{3}{2} = \frac{3}{4} = > \frac{3}{4} = m + 1 \\ m = \frac{3}{4} - 1 = > m = - \frac{1}{4} [/tex]
[tex]x^{2} - 2x - \frac{1}{4} + 1 = 0\\=>x^{2} - 2x +\frac{3}{4} = 0 <=> 4x^{2} - 8x + 3 = 0[/tex]
b)
[tex]x_{1}^{2} + x_{2}^{2} = x_{1}x_{2}(x_{1} + x_{2}) \\ x_{1}^{2} + x_{2}^{2} + 2x_{1}x_{2} - 2x_{1}x_{2} = x_{1}x_{2}(x_{1} + x_{2}) \\ {(x_{1} + x_{2})}^{2} - 2x_{1}x_{2} = x_{1}x_{2}(x_{1} + x_{2}) \\ {2}^{2} - 2(m + 1) = (m + 1) \times 2 \\ 4((m + 1) = 4 = > (m + 1) = 1 \\ = > m = 0[/tex]
[tex]x^{2} - 2x + m + 1 = 0\\=>x^{2} - 2x + 1 = 0[/tex]
