Question 33685
Well, I don't believe I've heard of the "diamond formula" for solving 4th-degree polynomials.
But try this method:
Solve:
{{{x^4 -5x^2+4 = 0}}} Rewrite this as:
{{{(x^2)^2 - 5(x^2) + 4 = 0}}} Temporarily set {{{x^2 = y}}}
{{{y^2 - 5y + 4 = 0}}} Solve this quadratic equation in y by factoring.
{{{(y - 1)(y - 4) = 0}}} Apply the zero products principle.
{{{y - 1 = 0}}} and/or {{{y - 4 = 0}}}
If {{{y - 1 = 0}}} then {{{y = 1}}}
If {{{y - 4 = 0}}} then {{{y = 4}}}

But, {{{y = x^2}}}, so:
{{{x^2 = 1}}} Take the square root of both sides.
{{{x = 1}}} or {{{x = -1}}} and:
{{{x^2 = 4}}} Take the square root of both sides.
{{{x = 2}}} or {{{x = -2}}}

The roots are:
x = 1
x = -1
x = 2
x = -2