Question 201889
{{{x^4-13x^2+36=0}}} Start with the given equation.



Let {{{z=x^2}}}. So {{{z^2=x^4}}}



{{{z^2-13z+36=0}}} Replace {{{x^4}}} with {{{z^2}}}. Replace {{{x^2}}} with "z"



Notice that the quadratic {{{z^2-13z+36}}} is in the form of {{{Az^2+Bz+C}}} where {{{A=1}}}, {{{B=-13}}}, and {{{C=36}}}



Let's use the quadratic formula to solve for "z":



{{{z = (-B +- sqrt( B^2-4AC ))/(2A)}}} Start with the quadratic formula



{{{z = (-(-13) +- sqrt( (-13)^2-4(1)(36) ))/(2(1))}}} Plug in  {{{A=1}}}, {{{B=-13}}}, and {{{C=36}}}



{{{z = (13 +- sqrt( (-13)^2-4(1)(36) ))/(2(1))}}} Negate {{{-13}}} to get {{{13}}}. 



{{{z = (13 +- sqrt( 169-4(1)(36) ))/(2(1))}}} Square {{{-13}}} to get {{{169}}}. 



{{{z = (13 +- sqrt( 169-144 ))/(2(1))}}} Multiply {{{4(1)(36)}}} to get {{{144}}}



{{{z = (13 +- sqrt( 25 ))/(2(1))}}} Subtract {{{144}}} from {{{169}}} to get {{{25}}}



{{{z = (13 +- sqrt( 25 ))/(2)}}} Multiply {{{2}}} and {{{1}}} to get {{{2}}}. 



{{{z = (13 +- 5)/(2)}}} Take the square root of {{{25}}} to get {{{5}}}. 



{{{z = (13 + 5)/(2)}}} or {{{z = (13 - 5)/(2)}}} Break up the expression. 



{{{z = (18)/(2)}}} or {{{z =  (8)/(2)}}} Combine like terms. 



{{{z = 9}}} or {{{z = 4}}} Simplify. 



So the solutions in terms of "z" are {{{z = 9}}} or {{{z = 4}}} 

  
  
Now recall that we let {{{z=x^2}}}



So when {{{z=9}}}, {{{x=sqrt(9)=3}}} or {{{x=-sqrt(9)=-3}}}



In other words, when {{{z=9}}}, {{{x=3}}} or {{{x=-3}}}



So the first two solutions in terms of "x" are: {{{x=3}}} or {{{x=-3}}}



Likewise, when {{{z=4}}}, {{{x=sqrt(4)=2}}} or {{{x=-sqrt(4)=-2}}}



So when {{{z=4}}}, {{{x=2}}} or {{{x=-2}}}



So the next two solutions in terms of "x" are: {{{x=2}}} or {{{x=-2}}}



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Answer:



So the four solutions are: 



{{{x=3}}}, {{{x=-3}}}, {{{x=2}}} or {{{x=-2}}}