Rozwiąż równanie a) 9x^ - 4 = 0b) x^ + 2x = 0c) 6x - 9 = x^d) 4x - 4x^ = 1e) x^ + V2x

rioter @rioter

October 2018 1 41 Report

Rozwiąż równanie a) 9x^ - 4 = 0

b) x^ + 2x = 0

c) 6x - 9 = x^

d) 4x - 4x^ = 1

e) x^ + V2x - 4 = 0

f) V3x^ - 9x + 6V3 = 0

Roma

$9x^2 - 4 = 0 \\\ (3x)^2 - 2^2 = 0 \\\ (3x - 2)(3x + 2) = 0 \\\ 3x - 2 = 0 \ \vee \ 3x + 2 = 0 \\\\ 3x - 2 = 0 \\\ 3x = 2 \ /: 3 \\\ x = \frac{2}{3} \\\\ 3x + 2 = 0 \\\ 3x = -2 \ /: 3 \\\ x = -\frac{2}{3}$

Odp. x = -⅔ i x = ⅔

$x^2+ 2x = 0 \\\ x \cdot (x + 2) = 0 \\\ x = 0 \ \vee \ x + 2 = 0 \\\\ x = 0 \\\\ x + 2 = 0 \\\ x = - 2$

Odp. x = - 2 i x = 0

$6x - 9 = x^2 \\\ -x^2 + 6x - 9 = 0 \ / \cdot (- 1) \\\ x^2 - 6x + 9 = 0 \\\ x^2 - 2 \cdot x \cdot 3 + 3^2 = 0 \\\ (x - 3)^2 = 0 \\\ x - 3 = 0 \\\ x = 3$

Odp. x = 3

$4x - 4x^2 = 1 \\\ - 4x^2 + 4x - 1 = 0 \ / \cdot (- 1)\\\ 4x^2 - 4x + 1 = 0 \\\ (2x)^2 - 2 \cdot 2x \cdot 1 + 1^2 = 0 \\\ (2x - 1)^2 = 0 \\\ 2x - 1 = 0 \\\ 2x = 1 \ /:2 \\\ x = \frac{1}{2}$

Odp. x = ½

$x^2 + \sqrt{2}x - 4 = 0 \\\ a = 1; \ b = \sqrt{2}; c = - 4 \\\ \Delta = b^2 - 4ac = (\sqrt{2})^2 - 4 \cdot 1 \cdot (-4) = 2 + 16 = 18\\\ \sqrt{\Delta} = \sqrt{18} = \sqrt{9 \cdot 2} = 3\sqrt{2} \\\\ x_1 = \frac{-b - \sqrt{\Delta}}{2a} = \frac{-\sqrt{2} - 3\sqrt{2}}{2 \cdot 1} = \frac{-4\sqrt{2}}{2} = - 2\sqrt{2} \\\\ x_2 = \frac{-b + \sqrt{\Delta}}{2a} = \frac{-\sqrt{2} + 3\sqrt{2}}{2 \cdot 1} = \frac{2\sqrt{2}}{2} = \sqrt{2}$

Odp. $x = -2\sqrt{2} \ i \ x = \sqrt{2}$

$\sqrt{3}x^2 - 9x + 6\sqrt{3} = 0 \\\ a = \sqrt{3}; \ b = -9; c =6\sqrt{3} \\\ \Delta = b^2 - 4ac = (-9)^2 - 4 \cdot \sqrt{3} \cdot 6\sqrt{3} = 81 - 72 = 9\\\ \sqrt{\Delta} = \sqrt{9} = 3$

$x_1 = \frac{-b- \sqrt{\Delta}}{2a} = \frac{9 - 3}{2 \cdot \sqrt{3}} = \frac{6}{2\sqrt{3}} = \frac{3}{\sqrt{3}} =\frac{3 \cdot \sqrt{3}}{\sqrt{3} \cdot \sqrt{3}} = \frac{3 \cdot \sqrt{3}}{3} = \sqrt{3} \\\\ x_2 = \frac{-b+ \sqrt{\Delta}}{2a} = \frac{9 + 3}{2 \cdot \sqrt{3}} = \frac{12}{2\sqrt{3}} = \frac{6}{\sqrt{3}} =\frac{6 \cdot \sqrt{3}}{\sqrt{3} \cdot \sqrt{3}} = \frac{6 \cdot \sqrt{3}}{3} = 2\sqrt{3}$