Question

Rezolvati in Recuatiile : 3xla puterea 2 +x-4=9 , 9x la puterea 2-9x+2=0 , (x+2)(2x-1)+x+4=0 vă rog frumos să mă ajutați!!!!!

Answer 1

$3 {x}^{2} + x - 4 = 9$

$3 {x}^{2} + x - 4 - 9 = 0$

$3 {x}^{2} + x - 13 = 0$

$a = 3$

$b = 1$

$c = - 13$

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

$\Delta = {1}^{2} - 4 \times 3 \times ( - 13)$

$\Delta = 1 + 156$

$\Delta = 157 >0 =>x_{1}\:\neq\:x_{2}$

$x_{1,2}=\frac{-b \: \pm \: \sqrt{\Delta}}{2a}$

$x_{1,2}=\frac{-1\pm\sqrt{157}}{2 \times 3} = \frac{ - 1 \pm \sqrt{157} }{6}$

$x_{1}=\frac{-1 + \sqrt{157}}{6}$

$x_{2}=\frac{ - 1 - \sqrt{157}}{6}$

$9 {x}^{2} - 9x + 2 = 0$

$a = 9$

$b = - 9$

$c = 2$

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

$\Delta = {( - 9)}^{2} - 4 \times 9 \times 2$

$\Delta = 81 - 72$

$\Delta = 9>0=>x_{1}\:\neq\:x_{2}$

$x_{1,2}=\frac{-b\pm\sqrt{\Delta}}{2a}$

$x_{1,2}=\frac{-( - 9)\pm\sqrt{9}}{2 \times 9}$

$x_{1,2}=\frac{9\pm3}{18}$

$x_{1}=\frac{9 + 3}{18} = \frac{12}{18} = \frac{2}{3}$

$x_{2}=\frac{9 - 3}{18} = \frac{6}{18} = \frac{1}{3}$

$(x + 2)(2x - 1) + x + 4 = 0$

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

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

$a = 2$

$b = 4$

$c = 2$

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

$\Delta = {4}^{2} - 4 \times 2 \times 2$

$\Delta = 16 - 16$

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

$x_{1} = x_{2} = \frac{ - b + \sqrt{\Delta} }{2a} = \frac{ - 4 + \sqrt{0} }{2 \times 2} = \frac{ - 4 + 0}{4} = - \frac{4}{4} = - 1$