X^-2+4x^-1+3=0 solve by making appropriate substitution

ANSWER[tex]x = - 1 \: or \: x = - \frac{1}{ 3} [/tex]EXPLANATIONThe given equation is:[tex] {x}^{ - 2} + 4 {x}^{ - 1} + 3 = 0[/tex]Recall that:[tex] {a}^{ - m} = \frac{1}{ {a}^{m} } [/tex][tex] \frac{1}{ {x}^{2} } + \frac{4}{x} + 3 = 0[/tex]Or[tex] {( \frac{1}{x} )}^{2} + 4( \frac{1}{x} ) + 3 = 0[/tex]Let [tex]u = \frac{1}{x} [/tex]Our equation then becomes:[tex] {u}^{2} + 4u + 3 = 0[/tex]The factors of 3 that add up to 4 are:[tex] {u}^{2} + 3u + u + 3[/tex][tex]u(u + 3) + 1(u + 3) = 0[/tex][tex](u + 1)(u + 3) = 0[/tex][tex]u + 1 = 0 \: or \: u + 3 = 0[/tex][tex]u = - 1 \: or \: u = - 3[/tex]This implies that:[tex] \frac{1}{x} = - 1 \: or \: \frac{1}{x} = - 3[/tex][tex]x = - 1 \: or \: x = - \frac{1}{ 3} [/tex]