Question 273250
9x^4-21x^2+10=0


Let k = x^2


Equation becomes:


9k^2 - 21k + 10 = 0


This equations factors out to be:


(3k-5) * (3k-2) = 0


solve for k to get:


k = 5/3 or k = 2/3


Since we set k = x^2, then:


k = 5/3 is the same as x^2 = 5/3
k = 2/3 is the same as x^2 = 2/3


Solving for x, we get:


x = +/- {{{sqrt(5/3)}}}
x = +/- {{{sqrt(2/3)}}}


plug these into your original equation to see if they hold up.


your original equation is:


9x^4 - 21x^2 + 10 = 0


I plugged these values into that equation and they all proved to be true so the values are good.


For example:


When x = {{{-sqrt(5/3)}}}, the equation becomes:


{{{9*(-sqrt(5/3))^4 - 21*(-sqrt(5/3))^2 + 10 = 0}}} which becomes:


25 - 35 + 10 = 0 which becomes:


-10 + 10 = 0 which becomes:


0 = 0 confirming that the value of {{{-sqrt(5/3)}}} is good.


Your answers are:


x = plus or minus {{{sqrt(5/3)}}}

and:


x = plus or minus {{{sqrt(2/3)}}}