Question

vă rog frumos repede ​

Answer 1

$a) \: {x}^{2} = 1 \\ x = \sqrt{1} \\ x = 1$

$b) \: {2x}^{2} + 1 = - 7 \\ {2x}^{2} = - 7 -1 \\ 2 {x}^{2} = - 8 \\ {x}^{2} = - 8 \div 2 \\ {x}^{2} = - 4 \\ x = - \sqrt{4} \\ x = 2$

$c) \: {x}^{2} + 2 {x}^{2} = {4x}^{2} \\ {3x}^{2} = {4x}^{2} \\ 3 {x}^{2} - {4x}^{2} = 0 \\ - {x}^{2} = 0 \\ {x}^{2} = 0 \\ x = 0$