Question 314663
Let {{{u=x^2}}}, then {{{u^2=x^4}}},
{{{36x^4 - 85x^2 + 49 = 0 }}}
{{{36u^2-85u+49=0}}}
You can factor,
{{{(36u-49)(u-1)=0}}}
Two "u" solutions:
{{{36u-49=0
{{{36u=49}}}
{{{u=49/36}}}
{{{x^2=49/36}}}
{{{x=0 +- 7/6}}}
.
.
.
{{{u-1=0}}}
{{{u=1}}}
{{{x^2=1}}}
{{{x=0 +- 1}}}
You then have 4 "x" solutions.