Question 180326
{{{9x^4-25x^2+16=0}}} Start with the given equation.


Let {{{u=x^2}}}. This means that {{{u^2=x^4}}}



{{{9u^2-25u+16=0}}} Replace {{{x^4}}} with {{{u^2}}} and {{{x^2}}} with {{{u}}}



Notice we have a quadratic equation in the form of {{{au^2+bu+c}}} where {{{a=9}}}, {{{b=-25}}}, and {{{c=16}}}



Let's use the quadratic formula to solve for u



{{{u = (-b +- sqrt( b^2-4ac ))/(2a)}}} Start with the quadratic formula



{{{u = (-(-25) +- sqrt( (-25)^2-4(9)(16) ))/(2(9))}}} Plug in  {{{a=9}}}, {{{b=-25}}}, and {{{c=16}}}



{{{u = (25 +- sqrt( (-25)^2-4(9)(16) ))/(2(9))}}} Negate {{{-25}}} to get {{{25}}}. 



{{{u = (25 +- sqrt( 625-4(9)(16) ))/(2(9))}}} Square {{{-25}}} to get {{{625}}}. 



{{{u = (25 +- sqrt( 625-576 ))/(2(9))}}} Multiply {{{4(9)(16)}}} to get {{{576}}}



{{{u = (25 +- sqrt( 49 ))/(2(9))}}} Subtract {{{576}}} from {{{625}}} to get {{{49}}}



{{{u = (25 +- sqrt( 49 ))/(18)}}} Multiply {{{2}}} and {{{9}}} to get {{{18}}}. 



{{{u = (25 +- 7)/(18)}}} Take the square root of {{{49}}} to get {{{7}}}. 



{{{u = (25 + 7)/(18)}}} or {{{u = (25 - 7)/(18)}}} Break up the expression. 



{{{u = (32)/(18)}}} or {{{u =  (18)/(18)}}} Combine like terms. 



{{{u = 16/9}}} or {{{u = 1}}} Simplify. 



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Now remember, we let {{{u=x^2}}}, so this means that 



{{{x^2 = 16/9}}} or {{{x^2 = 1}}}




Let's solve the first equation {{{x^2 = 16/9}}}:



{{{x = 0+-sqrt(16/9)}}} Take the square root of both sides



{{{x = sqrt(16/9)}}} or {{{x = -sqrt(16/9)}}} Break up the "plus/minus"



{{{x = 4/3}}} or {{{x = -4/3}}} Take the square root of {{{16/9}}} to get {{{4/3}}}




So the first two solutions are {{{x = 4/3}}} or {{{x = -4/3}}}



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Now let's solve the second equation {{{x^2 = 1}}}:



{{{x = 0+-sqrt(1)}}} Take the square root of both sides



{{{x = sqrt(1)}}} or {{{x = -sqrt(1)}}} Break up the "plus/minus"



{{{x = 1}}} or {{{x = -1}}} Take the square root of {{{1}}} to get {{{1}}}




So the next two solutions are {{{x = 1}}} or {{{x = -1}}}



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Answer:



So the 4 solutions are:



{{{x = 4/3}}}, {{{x = -4/3}}}, {{{x = 1}}} or {{{x = -1}}}