Question

X square + 1 whole square - x square =0 find the roots

Answer 1

Step-by-step explanation:

$0= ( {( {x}^{2}) + 1 )}^{2} - {x}^{2} \\ 0 = { (({x)}^{2} } + 1)^{2} - ( {x)}^{2} \\ 0 = ( {x}^{2} + 1 + x)( {x}^{2} + 1 - {x}^{2} ) \\ 0 = ( {x}^{2} + x + 1)(1) \\ 0= {x}^{2} + x + 1 \\ {x}^{2} + x + 1 = 0 \: \: \: match \: to \: \\ a {x}^{2} + bx + c = 0 \\ \: \\ a = 1 \: \: \: b = 1 \: \: \: c = 1 \\ use \: the \: formula \: of \: find \: roots \: of \: quadratic \: equation \: \\ \\ = \frac{ -b \frac{ + }{ - } \sqrt{ {b}^{2} - 4ac } }{2a} \\ = \frac{ - (1) \frac{ + }{ - } \sqrt{ {(1)}^{2} - 4(1)(1)} }{2(1)} \\ = \frac{ - 1 \frac{ + }{ - } \sqrt{1 - 4} }{2} \\ = \frac{ - 1 \frac{ + }{ - } \sqrt{3} }{2} \\ = \frac{ - 1 + \sqrt{3} }{2} \: \: \: \: \: \: or \: \: \: \: \: \: \frac{ - 1 - \sqrt{3} }{2} \\ \: the \: roots \: of \: equation \: is \: \\ \frac{ - 1 + \sqrt{3} }{2} \: \: \: and \: \: \: \: \frac{ - 1 - \sqrt{3} }{2}$