Question 158768
i got {{{x=+sqrt(5/2)}}} and {{{x=-sqrt(5/2)}}}
i also got {{{x=+sqrt(3/2)}}} and {{{x=-sqrt(3/2)}}}
here's how.....
{{{4*x^4 - 16*x + 15 = (2x^2-5)*(2x^2-3) = 0}}}
if the whole thing equals 0 then at least one of the factors must = 0.
solving for the first factor = 0, i get
{{{(2*x^2-5) = 0}}} becomes {{{(2x^2)=5}}} becomes {{{x^2=(5/2)}}}
and
solving for the second factor = 0, i get
{{{(2*x^2-3) = 0}}} becomes {{{(2x^2)=3}}} becomes {{{x^2=(3/2)}}}
plugging {{{x^2=(5/2)}}} into the original equation, i get
{{{4*(5/2)^2 - 16*(5/2) + 15 = 0}}}
this becomes
{{{4*(25/4) - 80/2 + 15 = 0}}}
this becomes {{{-15+15=0}}} proving first factor is good.
plugging {{{x^2=(3/2)}}} into the original equation, i get
{{{4*(3/2)^2 - 16*(3/2) + 15 = 0}}}
this becomes
{{{4*(9/4) - 48/2 + 15 = 0}}}
this becomes {{{-15+15=0}}} proving second factor is good.
if {{{x^2 = (5/2)}}} and {{{x^2 = (3/2)}}} are the correct answers (proven above), then
x = +/- {{{sqrt(5/2)}}} and x = +/- {{{sqrt(3/2)}}} are the correct values for x.