Zadanie w załączniku :)
a)
[tex]49x^{2}-4 = 0\\\\(7x+2)(7x-2) = 0\\\\7x+2 = 0 \ \vee \ 7x-2 = 0\\\\7x = -2 \ \vee \ 7x = 2\\\\x = -\frac{2}{7} \ \ \vee \ \ x = \frac{2}{7}\\\\\boxed{x \in\left\{-\frac{2}{7},\frac{2}{7}\right\}}[/tex]
b)
[tex]-2x^{2}=3x\\\\-2x^{2}-3x = 0 \ \ \ |\cdot(-1)\\\\2x^{2}+3x = 0\\\\x(2x+3) = 0\\\\x = 0 \ \vee \ 2x+3 = 0\\\\x = 0 \ \vee \ 2x = -3\\\\x = 0 \ \vee \ x = -\frac{3}{2}\\\\\boxed{x \in\left\{-\frac{3}{2},0\right\}}[/tex]
c)
[tex]9x^{2}=12x-4\\\\9x^{2}-12x+4 =0\\\\(3x-2)^{2} = 0\\\\3x-2 = 0\\\\3x = 2 \ \ \ /:3\\\\\boxed{x_{o}= \frac{2}{3}}[/tex]
" Life is not a problem to be solved but a reality to be experienced! "
© Copyright 2013 - 2025 KUDO.TIPS - All rights reserved.
Verified answer
Zadanie w załączniku :)
a)
[tex]49x^{2}-4 = 0\\\\(7x+2)(7x-2) = 0\\\\7x+2 = 0 \ \vee \ 7x-2 = 0\\\\7x = -2 \ \vee \ 7x = 2\\\\x = -\frac{2}{7} \ \ \vee \ \ x = \frac{2}{7}\\\\\boxed{x \in\left\{-\frac{2}{7},\frac{2}{7}\right\}}[/tex]
b)
[tex]-2x^{2}=3x\\\\-2x^{2}-3x = 0 \ \ \ |\cdot(-1)\\\\2x^{2}+3x = 0\\\\x(2x+3) = 0\\\\x = 0 \ \vee \ 2x+3 = 0\\\\x = 0 \ \vee \ 2x = -3\\\\x = 0 \ \vee \ x = -\frac{3}{2}\\\\\boxed{x \in\left\{-\frac{3}{2},0\right\}}[/tex]
c)
[tex]9x^{2}=12x-4\\\\9x^{2}-12x+4 =0\\\\(3x-2)^{2} = 0\\\\3x-2 = 0\\\\3x = 2 \ \ \ /:3\\\\\boxed{x_{o}= \frac{2}{3}}[/tex]