Answer
$$ - 3;3; - \sqrt 2 ;\sqrt 2 $$
Work Step by Step
$$\eqalign{
& {u^4} - 11{u^2} + 18 = 0 \cr
& {\left( {{u^2}} \right)^2} - 11\left( {{u^2}} \right) + 18 = 0 \cr
& {\text{we can set }}x = {u^2}.{\text{ then}} \cr
& {\left( x \right)^2} - 11\left( x \right) + 18 = 0 \cr
& {x^2} - 11x + 18 = \cr
& {\text{factoring the trinomial}} \cr
& \left( {x - 9} \right)\left( {x - 2} \right) = 0 \cr
& {\text{use zero - product property}} \cr
& x - 9 = 0{\text{ or }}x - 2 = 0 \cr
& x = 9\,\,\,\,\,{\text{or }}\,\,\,x = 2 \cr
& {\text{replace }}{u^2}{\text{ for }}x \cr
& {u^2} = 9\,\,\,\,\,{\text{or }}\,\,\,{u^2} = 2 \cr
& {\text{then}} \cr
& u = \pm \sqrt 9 {\text{ or }}u = \pm \sqrt 2 \cr
& u = \pm 3{\text{ or }}u = \pm \sqrt 2 \cr
& {\text{the solutions are}}: - 3;3; - \sqrt 2 ;\sqrt 2 \cr} $$
