Question

find x,,,,,,pls it's urgent​

Answer 1

Answer:

$( {x}^{2} + 1)^{2} - 5 {x}^{2} - 5 = 0 \\ (x ^{2} )^{2} + 2( {x}^{2} )1 + {1}^{2} - 5 {x}^{2} - 5 = 0 \\ {x}^{4} + 2 {x}^{2} + 1 - 5 {x}^{2} - 5 = 0\\ {x}^{4} - 3 {x}^{2} - 4 = 0 \\ {x}^{4} - 4 {x}^{2} + {x}^{2} - 4 = 0 \\ {x}^{2} ( {x}^{2} - 4) + 1( {x}^{2} - 4) = 0 \\ ({x}^{2} - 4)( {x}^{2} + 1) = 0 \\ {x}^{2} - 4 = 0 \: \: or \: \: {x}^{2} + 1 = 0 \\ {x}^{2} = 4 \: \: or \: \: {x}^{2} = - 1 \\ x = \sqrt{4} \: \: or \: \: x = \sqrt{ - 1} \\ x = 2 \: \: or \: \: x = \sqrt{ - 1}$