Answer
$$x \approx - 1.249621068$$
Work Step by Step
$$\eqalign{
& {x^4} + {x^2} - 4 = 0;{\text{ }}x < 0 \cr
& {\text{From the graph we can see that the equation has 2 solutions}} \cr
& {\text{Let }}f\left( x \right) = {x^4} + {x^2} - 4 \cr
& f'\left( x \right) = 4{x^3} + 2x \cr
& {\text{Using the Newton's Method }} \cr
& {x_{n + 1}} = {x_n} - \frac{{f\left( {{x_n}} \right)}}{{f'\left( {{x_n}} \right)}},{\text{ }}n = 1,2,3,...{\text{ }}\left( {\bf{1}} \right) \cr
& {\text{Substituting }}f\left( x \right){\text{ and }}f'\left( x \right){\text{ into }}\left( {\bf{1}} \right) \cr
& {x_{n + 1}} = {x_n} - \frac{{x_n^4 + x_n^2 - 4}}{{4x_n^3 + 2{x_n}}}{\text{ }}\left( {\bf{2}} \right) \cr
& {\text{From the graph we can see that the possible initial }} \cr
& {\text{approximation is }}{x_1} = - 1,{\text{ substituting into }}\left( {\bf{2}} \right) \cr
& {x_2} = \left( { - 1} \right) - \frac{{{{\left( { - 1} \right)}^4} + {{\left( { - 1} \right)}^2} - 4}}{{4{{\left( { - 1} \right)}^3} + 2\left( { - 1} \right)}} \cr
& {x_2} \approx - 1.333333333 \cr
& {\text{Continuing the process we obtain}} \cr
& {x_3} \approx - 1.256097561 \cr
& {x_4} \approx - 1.249662991 \cr
& {x_5} \approx - 1.219621069 \cr
& {x_6} \approx - 1.249621068 \cr
& {x_7} \approx - 1.249621068 \cr
& {x_6} = {x_7}{\text{ into 9 decimal places}}{\text{, then the solution for }}x < 0{\text{ of}} \cr
& {\text{the equation is }} \cr
& x \approx - 1.249621068 \cr} $$
