HELP ME! найдите корни уравнения: 1) - 3x^2 - 4x = 5 + 2x^2 - 4x 2) 2x^2 - 14 + 3x=3x 3) 2x^2 - 4x = -(4x-3). Created by SlivkO0O154. algebra-ru

HELP ME! найдите корни уравнения: 1) - 3x^2 - 4x = 5 + 2x^2 - 4x 2) 2x^2

SlivkO0O154 @SlivkO0O154

July 2022 1 5 Report

HELP ME!
найдите корни уравнения:
1) - 3x^2 - 4x = 5 + 2x^2 - 4x
2) 2x^2 - 14 + 3x=3x
3) 2x^2 - 4x = -(4x-3)

Answers & Comments

Daliki008

Ответ:

$1) - 3 {x}^{2} - 4x = 5 + 2 {x}^{2} - 4x \\ 0 = 5 + 2 {x}^{2} - 4x + 4x + 3 {x}^{2} \\ 0 = 5 + 5 {x}^{2} \\ 5 {x}^{2} + 5 = 0 \\ 5 {x}^{2} = - 5 \\ {x}^{2} = \frac{ - 5}{5} \\ x = + - \sqrt{ - 1} \\ non$

$2)2 {x}^{2} - 14 + 3x = 3x \\ 2 {x}^{2} - 14 + 3x - 3x = 0 \\ 2 {x}^{2} - 14 = 0 \\ 2 {x}^{2} = 14 \\ {x}^{2} = \frac{14}{2} \\ x1 = - \sqrt{7} \\ x2 = + \sqrt{7}$

$3)2 {x}^{2} - 4x = - (4x - 3) \\ 2 {x}^{2} - 4x = - 4x + 3 \\ 2 {x}^{2} - 4x + 4x - 3 = 0 \\ 2 {x}^{2} - 3 = 0 \\ {x}^{2} = \frac{3}{2} \\ x1 = - \sqrt{ \frac{3}{2} } \\ x2 = + \sqrt{ \frac{3}{2} }$

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