Question 252245
solving the equation:
{{{x^4 - 5 x^2 + 6 = 0}}}
Let {{{z = x^2}}}
{{{ z^2 - 5z + 6 = 0 }}}
{{{z = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
{{{a = 1}}}
{{{b = -5}}}
{{{c = 6}}}
{{{z = (-(-5) +- sqrt( (-5)^2-4*1*6 ))/(2*1) }}}
{{{z = (5 +- sqrt(25 - 24 ))/2 }}}
{{{z = (5 + 1)/2}}}
{{{z = 3}}}
and
{{{z = (5 - 1)/2}}}
{{{z = 2}}}
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Substituting:
{{{x^2 = 3}}}
{{{x = sqrt(3)}}}
{{{x = -sqrt(3)}}}
{{{x^2 = 2}}}
{{{x = sqrt(2)}}}
{{{x = -sqrt(2)}}}
The sum of the squares of the roots is:
{{{3 + 3 + 2 + 2 = 10}}}