Question

cum rezolv ecuatiile:
$a){x }^{2} + 6x = 16 \\ b) {x}^{2} - 2x = 8 \\ c)4 {x}^{2} - 4x = 12$

Answer 1

$a) {x}^{2} + 6x = 16$

${x}^{2} + 6x - 16 = 0$

${x}^{2} + 8x - 2x - 16 = 0$

$x(x + 8) - 2(x + 8) = 0$

$(x + 8)(x - 2) = 0$

$x +8 = 0 = > x_{1} = -8$

$x - 2 = 0 = > x_{2} = 2$

$b) {x}^{2} - 2x = 8$

${x}^{2} - 2x - 8 = 0$

${x}^{2} - 4x + 2x - 8 = 0$

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

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

$x - 4 = 0 = > x_{1} = 4$

$x + 2 = 0 = > x_{2} = - 2$

$c)4{x}^{2} - 4x = 12$

$4 {x}^{2} - 4x - 12 = 0 \: | \div 4$

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

$a = 1$

$b = - 1$

$c = - 3$

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

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

$\Delta = 1 + 12$

$\Delta = 13$

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

$x_{1,2}=\frac{-( - 1)\pm\sqrt{13}}{2 \times 1}$

$x_{1,2}=\frac{1\pm\sqrt{13}}{2}$

$x_{1}=\frac{1 + \sqrt{13}}{2}$

$x_{2}=\frac{1 - \sqrt{13}}{2}$