Let us first factorize #x^4-10x^2+9#.

#x^4-10x^2+9=x^4-9x^2-x^2+9#

= #x^2(x^2-9)-1(x^2-9)= (x^2-9)(x^2-1))#

= #(x+3)(x-3)(x+1)(x-1)#

Hence we have to solve the inequality

#(x+3)(x-3)(x+1)(x-1)>=0#

From this we know that the product #(x+3)(x-3)(x+1)(x-1)# has to be zero or positive. It is apparent that sign of binomials #(x+3)#, #(x+1)#, #(x-1)# and #(x-3)# will change around the values #-3#. #-1#, #1# and #3# respectively. In sign chart we divide the real number line using these values, i.e. below #-3#, between #-3# and #-1#, between #-1# and #1#, between #1# and #3# and above #3# and see how the sign of #(x+3)(x-3)(x+1)(x-1)# changes.

**Sign Chart**

#color(white)(XXXXXXXX)-3color(white)(XXXX)-1color(white)(XXXX)1color(white)(XXXX)3#

#(x+3)color(white)(XX)-ive color(white)(XX)+ive color(white)(XXX)+ive color(white)(XX)+ive color(white)(XX)+ive#

#(x+1)color(white)(XX)-ive color(white)(XX)-ive color(white)(XXX)+ive color(white)(XX)+ive color(white)(XX)+ive#

#(x-1)color(white)(XX)-ive color(white)(XX)-ive color(white)(XXX)-ive color(white)(XX)+ive color(white)(XX)+ive#

#(x-3)color(white)(XX)-ive color(white)(XX)-ive color(white)(XXX)-ive color(white)(XX)-ive color(white)(XX)+ive#

#x^4-10x^2+9#

#color(white)(XXXXXX)+ive color(white)(Xx)-ive color(white)(XXX)+ive color(white)(XX)-ive color(white)(XX)+ive#

It is observed that #x^4-10x^2+9 >= 0#

when either #-oo <= x <= -3# or #-1 <= x <= 1#, or #3 <= x <= oo#, which is the solution for the inequality.

In interval notation, this can be written as #[-oo,-3]uu[-1,1]uu[3,oo]#