Question 142982
{{{y^4-3y^2=-1}}} Start with the given equation



{{{y^4-3y^2+1=0}}} Add 1 to both sides



Let {{{x=y^2}}}. So this means that {{{x^2=y^4}}}



{{{x^2-3x+1=0}}} Replace {{{y^4}}} with {{{x^2}}}. Replace {{{y^2}}} with {{{x}}}. 



Let's use the quadratic formula to solve for x:



Starting with the general quadratic


{{{ax^2+bx+c=0}}}


the general solution using the quadratic equation is:


{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a)}}}




So lets solve {{{x^2-3*x+1=0}}} ( notice {{{a=1}}}, {{{b=-3}}}, and {{{c=1}}})





{{{x = (--3 +- sqrt( (-3)^2-4*1*1 ))/(2*1)}}} Plug in a=1, b=-3, and c=1




{{{x = (3 +- sqrt( (-3)^2-4*1*1 ))/(2*1)}}} Negate -3 to get 3




{{{x = (3 +- sqrt( 9-4*1*1 ))/(2*1)}}} Square -3 to get 9  (note: remember when you square -3, you must square the negative as well. This is because {{{(-3)^2=-3*-3=9}}}.)




{{{x = (3 +- sqrt( 9+-4 ))/(2*1)}}} Multiply {{{-4*1*1}}} to get {{{-4}}}




{{{x = (3 +- sqrt( 5 ))/(2*1)}}} Combine like terms in the radicand (everything under the square root)




{{{x = (3 +- sqrt(5))/(2*1)}}} Simplify the square root (note: If you need help with simplifying the square root, check out this <a href=http://www.algebra.com/algebra/homework/Radicals/simplifying-square-roots.solver> solver</a>)




{{{x = (3 +- sqrt(5))/2}}} Multiply 2 and 1 to get 2


So now the expression breaks down into two parts


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




So the solutions in terms of x are




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




But remember, we let {{{x=y^2}}}. So this means 



{{{y^2 = (3 + sqrt(5))/2}}} or {{{y^2 = (3 - sqrt(5))/2}}}



Take the square root of both sides in each case


{{{y =0+-sqrt( (3 + sqrt(5))/2)}}} or {{{y =0+-sqrt( (3 - sqrt(5))/2)}}}



So after breaking up the expressions, we get the four solutions


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