Question

Doresc va rog frumos o rezolvare pentru acest exercitiu : ( clasa a 9 ) Matematica :
1 ) Se considera ecuatia x^2-2x+m+1=0 , cu radacinile x 1 , x 2 . Determinati ecuatiile
reale , in cazurile :
a ) x 1 = 3 x 2
b ) x 1 ^2 + x 2 ^2 = x 1 x 2 (x 1 + x 2) .

Answer 1

Explicație pas cu pas:

$x^{2} - 2x + m + 1 = 0$

$a = 1; \: b = -2 ; \: c = m + 1$

$Δ \geqslant 0 = > ( - 2)^{2} - 4(m + 1) \geqslant 0 \\ 4 - 4m - 4 \geqslant 0 \\ 4m \leqslant 0 = > m \leqslant 0$

$x_{1} + x_{2} = \frac{ - b}{a} = \frac{ - ( - 2)}{1} = 2$

$x_{1}x_{2} = \frac{c}{a} = \frac{m + 1}{1} = m + 1$

a)

$x_{1} = 3x_{2}$

$3x_{2} + x_{2} = 2 \\ 4x_{2} = 2 = > x_{2} = \frac{1}{2}$

$x_{1} = 3 \times \frac{1}{2} = \frac{3}{2}$

$x_{1}x_{2} = \frac{1}{2} \times \frac{3}{2} = \frac{3}{4} = > \frac{3}{4} = m + 1 \\ m = \frac{3}{4} - 1 = > m = - \frac{1}{4}$

$x^{2} - 2x - \frac{1}{4} + 1 = 0\\=>x^{2} - 2x +\frac{3}{4} = 0 <=> 4x^{2} - 8x + 3 = 0$

b)

$x_{1}^{2} + x_{2}^{2} = x_{1}x_{2}(x_{1} + x_{2}) \\ x_{1}^{2} + x_{2}^{2} + 2x_{1}x_{2} - 2x_{1}x_{2} = x_{1}x_{2}(x_{1} + x_{2}) \\ {(x_{1} + x_{2})}^{2} - 2x_{1}x_{2} = x_{1}x_{2}(x_{1} + x_{2}) \\ {2}^{2} - 2(m + 1) = (m + 1) \times 2 \\ 4((m + 1) = 4 = > (m + 1) = 1 \\ = > m = 0$

$x^{2} - 2x + m + 1 = 0\\=>x^{2} - 2x + 1 = 0$