👤

X la puterea a doua +x+1=5x-2as dori rezolvarea si cum se trec termeni in partea cealalta

Răspuns :

[tex] {x}^{2} + x + 1 = 5x - 2[/tex]

[tex] {x}^{2} + x + 1 - 5x + 2 = 0[/tex]

[tex] {x}^{2} + x - 5x + 1 + 2 = 0[/tex]

[tex] {x}^{2} - 4x + 3 = 0[/tex]

M1)

[tex] {x}^{2} - 3x - x + 3 = 0[/tex]

[tex]x(x - 3) - (x - 3) = 0[/tex]

[tex](x - 3)(x - 1) = 0[/tex]

[tex]x - 3 = 0 = > x = 3[/tex]

[tex]x - 1 = 0 = > x = 1[/tex]

M2)

[tex] {x}^{2} - 4x + 3 = 0[/tex]

[tex]a = 1[/tex]

[tex]b = - 4[/tex]

[tex]c = 3[/tex]

[tex]\Delta = {b}^{2} - 4ac[/tex]

[tex]\Delta = {( - 4)}^{2} - 4 \times 1 \times 3[/tex]

[tex]\Delta = 16 - 12[/tex]

[tex]\Delta = 4[/tex]

[tex]x_{1,2}=\frac{-b\pm\sqrt{\Delta}}{2a} = \frac{ - ( - 4) \pm \sqrt{4} }{2 \times 1} = \frac{4 \pm2}{2} [/tex]

[tex]x_{1}= \frac{4 + 2}{2} = \frac{6}{2} = 3[/tex]

[tex]x_{2}= \frac{4 - 2}{2} = \frac{2}{2} = 1[/tex]