Question

Va rog frumos exercitiile 3 si 4!

Answer 1

$3)a)3 {x}^{2} = 70$

${x}^{2} = \frac{70}{3}$

$x = \pm \sqrt{ \frac{70}{3} }$

$x = \pm \frac{ \sqrt{70} }{ \sqrt{3} }$

$x = \pm \frac{ \sqrt{210} }{3}$

$b) \frac{x}{2} = \frac{8}{x}$

${x}^{2} = 16$

$x = \pm \sqrt{16}$

$x = \pm4$

$c) \frac{3x}{5} = \frac{20}{3x}$

$9 {x}^{2} = 100$

${x}^{2} = \frac{100}{9}$

$x = \pm \sqrt{ \frac{100}{9} }$

$x = \pm \frac{ \sqrt{100} }{ \sqrt{9} }$

$x = \pm \frac{10}{3}$

$d) {(x + 3)}^{2} = 25$

$x + 3 = \pm \sqrt{25}$

$x + 3 = \pm5$

$1)x + 3 = 5 = > x = 5 - 3 = 2$

$2)x + 3 = - 5 = > x = - 5 - 3 = - 8$

$e) {(3x - 1)}^{2} = 49$

$3x - 1 = \pm \sqrt{49}$

$3x - 1 = \pm7$

$1)3x - 1 = 7 = > 3x = 7 + 1 = > 3x = 8 = > x = \frac{8}{3}$

$2)3x - 1 = - 7 = > 3x = - 7 + 1 = > 3x = - 6 = > x = - \frac{6}{3} = > x = - 2$

$f) {x}^{2} + 4x + 4 = 16$

${x}^{2} + 4x + 4 - 16 = 0$

${x}^{2} + 4x - 12 = 0$

${x}^{2} + 6x - 2x - 12 = 0$

$x(x + 6) - 2(x + 6) = 0$

$(x + 6)(x - 2) = 0$

$1)x + 6 = 0 = > x = - 6$

$2)x - 2 = 0 = > x = 2$

$g) {x}^{2} + 25 = 10x + 25$

${x}^{2} = 10x \: | \div x$

$x = 10$

$h) {x}^{2} + 25 = 10x + 1$

${x}^{2} - 10x + 25 - 1 = 0$

${x}^{2} - 10x + 24 = 0$

${x}^{2} - 6x - 4x + 24 = 0$

$x(x - 6) - 4(x - 6) = 0$

$(x - 6)(x - 4) = 0$

$1)x - 6 = 0 = > x = 6$

$2)x - 4 = 0 = > x = 4$

$i) {x}^{2} + 8x + 15 = 0$

${x}^{2} + 3x + 5x + 15 = 0$

$x(x + 3) + 5(x + 3) = 0$

$(x + 3)(x + 5) = 0$

$1)x + 3 = 0 = > x = - 3$

$2)x + 5 = 0 = > x = - 5$

$j) {x}^{4} = 81$

${x}^{4} = {3}^{4}$

$x = 3$

$k) {x}^{2} + 3x + 2 = 0$

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

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

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

$1)x + 2 = 0 = > x = - 2$

$2)x + 1 = 0 = > x = - 1$

$l) {x}^{2} - 2x + 3 = 0$

$nu \: are \: radacini \: reale$

$4)a) {x}^{2} - 1 = 24$

${x}^{2} = 24 + 1$

${x}^{2} = 25$

$x = \pm \sqrt{25}$

$x = \pm5$

$b)4 {x}^{2} + 4x + 1 = 9$

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

$\Delta = 16 - 4 \times 4 \times ( - 8) = 16 + 128 = 144$

$x_{1} = \frac{ - 4 + 12}{8} = 1$

$x_{2} = \frac{ - 4 - 12}{8} = - 2$

$c) {(5x - 1)}^{2} = 36$

$5x - 1 = \pm \sqrt{36}$

$5x - 1 = \pm6$

$1)5x- 1 = 6 = > 5x = 7 = > x = \frac{7}{5}$

$2)5x - 1 = - 6 = > 5x = - 5 = > x = - 1$

$d) {(2x - 3)}^{2} = 4$

$2x - 3 = \pm \sqrt{4}$

$2x - 3 = \pm2$

$1)2x - 3 = 2 = > 2x = 5 = > x = \frac{5}{2}$

$2)2x - 3 = - 2 = > 2x = 1 = > x = \frac{1}{2}$

$e) {(2 - 3x)}^{2} = - 4$

$2 - 3x = \pm \sqrt{ - 4} = > ecuatia \: nu \: are \: radacini \: reale$

$f)25 {x}^{2} = 4$

${x}^{2} = \frac{4}{25}$

$x = \pm \sqrt{ \frac{4}{25} }$

$x = \pm \frac{ \sqrt{4} }{ \sqrt{25} }$

$x = \pm \frac{2}{5}$

$g)3 {x}^{2} = 27$

${x}^{2} = \frac{27}{3}$

${x}^{2} = 9$

$x = \pm \sqrt{9}$

$x = \pm3$

$h) {x}^{2} = 11 - 6 \sqrt{2}$

$x = \pm \sqrt{11 - 6 \sqrt{2} }$

$i) {x}^{2} + 7 = 0$

${x}^{2} = - 7$

$x = \pm \sqrt{ - 7} = > ecuatia \: nu \: are \: radacini \: reale$

$j) \frac{x}{3} = \frac{3}{16x}$

$16 {x}^{2} = 9$

${x}^{2} = \frac{9}{16}$

$x = \pm \sqrt{ \frac{9}{16} }$

$x = \pm \frac{ \sqrt{9} }{ \sqrt{16} }$

$x = \pm \frac{3}{4}$

$k) \frac{9}{25x} = \frac{x}{49}$

$25 {x}^{2} = 441$

${x}^{2} = \frac{441}{25}$

$x = \pm \sqrt{ \frac{441}{25} }$

$x = \pm \frac{ \sqrt{441} }{ \sqrt{25} }$

$x = \pm \frac{21}{5}$

$l) \frac{5}{ {x}^{2} } = \frac{64}{5}$

$64 {x}^{2} = 25$

${x}^{2} = \frac{25}{64}$

$x = \pm \sqrt{ \frac{25}{64} }$

$x = \pm \frac{ \sqrt{25} }{ \sqrt{64} }$

$x = \pm \frac{5}{8}$