Question

Calculati:a)(x-1)^2=1 b)x^2-1=1 c)x^2+6x+9=16 d)(x+2)^2-4=0 e)2x+3 supra 3=9 supra 2x-3

Answer 1

$a) {(x - 1)}^{2} = 1$

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

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

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

$x(x - 2) = 0$

$= > x_{1} = 0$

$x - 2 = 0$

$= > x_{2} = 2$

$b) {x}^{2} - 1 = 1$

${x}^{2} = 1 + 1$

${x}^{2} = 2$

$x = \pm \sqrt{2}$

$x_{1}=\sqrt{2}$

$x_{2}=-\sqrt{2}$

$c) {x}^{2} + 6x + 9 = 16$

${(x + 3)}^{2} = 16$

$x + 3 = \pm \sqrt{16}$

$x+3=\pm4$

$1)x + 3 = 4$

$x = 4 - 3$

$x_{1} = 1$

$2)x + 3 = -4$

$x = - 4 - 3$

$x_{2} = - 7$

$d) {(x + 2)}^{2} - 4 = 0$

${(x + 2)}^{2} = 4$

$x + 2 = \pm \sqrt{4}$

$x + 2 = \pm2$

$1)x + 2 = 2$

$x = 2 - 2$

$x_{1} = 0$

$2)x + 2 = - 2$

$x = - 2 - 2$

$x_{2} = - 4$

$e) \frac{2x + 3}{3} = \frac{9}{2x - 3}$

$(2x + 3)(2x - 3) = 3 \times 9$

${4x}^{2} - 9 = 27$

$4 {x}^{2} = 27 + 9$

$4 {x}^{2} = 36 \: | \div 4$

${x}^{2} = 9$

$x = \pm \sqrt{9}$

$x = \pm3$

$x_{1} = 3$

$x_{2} = - 3$