Question

radical mare din 2x^2-6x+5 =x-1

Answer 1

Sper ca te-am ajutat. Mult succes mai departe.

Answer 2

$\sqrt{2 {x}^{2} - 6x + 5} = x - 1 \: |{( \: )}^{2}$

$2 {x}^{2} - 6x + 5 \geqslant 0$

$x - 1 \geqslant 0 = > x \geqslant 1$

$2 {x}^{2} - 6x + 5 = {(x - 1)}^{2}$

$2 {x}^{2} - 6x + 5 = {x}^{2} - 2x + 1$

$2 {x}^{2} - {x}^{2} - 6x + 2x + 5 - 1 = 0$

${x}^{2} - 4x + 4 = 0$

$a = 1$

$b = - 4$

$c = 4$

$\Delta = {b}^{2} - 4ac$

$\Delta = {( - 4)}^{2} - 4 \times 1 \times 4$

$\Delta = 16 - 16 = 0 = > \exists \: x_{1} = \: x_{2}$

$x_{1} = \frac{ - b + \sqrt{\Delta} }{2a} = \frac{ - ( - 4) + \sqrt{0} }{2 \times 1} = \frac{4 + 0}{2} = \frac{4}{2} = 2 \: verifica \: conditiile$