Question 175567
{{{x^4 + 2x^2-1 = 0}}} 
You can use a substitution and knock this down to a quadratic equation.
Let {{{z=x^2}}}
{{{x^4+2x^2-1=0}}}
{{{z^2+2z-1=0}}}
Use the quadratic formula,
{{{z = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
{{{z = (-2 +- sqrt( 2^2-4*1*(-1)))/(2*1) }}}
{{{z = (-2 +- sqrt( 4+4))/(2) }}}
{{{z = -1 +- sqrt(2)}}}
Two solutions for z.
.
.
First solution for z,
{{{z = -1 + sqrt(2)}}}
{{{x^2 = -1 + sqrt(2)}}}
Two real roots,
{{{x =0 +- sqrt(-1 + sqrt(2))}}}
or approximately
{{{x=0.643}}} and {{{x=-0.643}}}
.
.
Second solution for z,
{{{z = -1 - sqrt(2)}}}
{{{x^2= -1 - sqrt(2)}}}
Two complex roots,
{{{x^2= (-1)*(1+sqrt(2))}}}
{{{x= 0 +- sqrt((-1)*(1+sqrt(2)))}}}
{{{x= 0 +- sqrt(1+sqrt(2))i}}}
or approximately
{{{x=1.554i}}} and {{{x=-1.554i}}}
.
.
.
The graph verifies the real roots.
.
.
.
{{{ graph( 300, 300, -2, 2, -10, 10, x^4+2x^2-1) }}}