Question 480750
{{{15x^4-28x^2+5=0}}} Start with the given equation.



Let {{{z=x^2}}}. So this means that {{{z^2=(x^2)^2=x^4}}}



{{{15z^2-28z+5=0}}} Replace {{{x^4}}} with {{{z^2}}} and {{{x^2}}} with "z"



Notice we have a quadratic equation in the form of {{{az^2+bz+c}}} where {{{a=15}}}, {{{b=-28}}}, and {{{c=5}}}



Let's use the quadratic formula to solve for z



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



{{{z = (-(-28) +- sqrt( (-28)^2-4(15)(5) ))/(2(15))}}} Plug in  {{{a=15}}}, {{{b=-28}}}, and {{{c=5}}}



{{{z = (28 +- sqrt( (-28)^2-4(15)(5) ))/(2(15))}}} Negate {{{-28}}} to get {{{28}}}. 



{{{z = (28 +- sqrt( 784-4(15)(5) ))/(2(15))}}} Square {{{-28}}} to get {{{784}}}. 



{{{z = (28 +- sqrt( 784-300 ))/(2(15))}}} Multiply {{{4(15)(5)}}} to get {{{300}}}



{{{z = (28 +- sqrt( 484 ))/(2(15))}}} Subtract {{{300}}} from {{{784}}} to get {{{484}}}



{{{z = (28 +- sqrt( 484 ))/(30)}}} Multiply {{{2}}} and {{{15}}} to get {{{30}}}. 



{{{z = (28 +- 22)/(30)}}} Take the square root of {{{484}}} to get {{{22}}}. 



{{{z = (28 + 22)/(30)}}} or {{{z = (28 - 22)/(30)}}} Break up the expression. 



{{{z = (50)/(30)}}} or {{{z =  (6)/(30)}}} Combine like terms. 



{{{z = 5/3}}} or {{{z = 1/5}}} Simplify. 



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Since we let {{{z=x^2}}}, we need to plug that in:



{{{z = 5/3}}} Start with the first solution for "z"



{{{x^2 = 5/3}}} Plug in {{{z=x^2}}}



{{{x = 0+-sqrt(5/3)}}} Take the square root of both sides.



{{{x = sqrt(5/3)}}} or {{{x = -sqrt(5/3)}}} Break up the "plus/minus".



{{{x = sqrt(15)/3}}} or {{{x = -sqrt(15)/3}}} Simplify the square root.



So the first two solutions are {{{x = sqrt(15)/3}}} or {{{x = -sqrt(15)/3}}}



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{{{z = 1/5}}} Start with the second solution for "z"



{{{x^2 = 1/5}}} Plug in {{{z=x^2}}}



{{{x = 0+-sqrt(1/5)}}} Take the square root of both sides.



{{{x = sqrt(1/5)}}} or {{{x = -sqrt(1/5)}}} Break up the "plus/minus".



{{{x = sqrt(5)/5}}} or {{{x = -sqrt(5)/5}}} Simplify the square root.



So the next two solutions are {{{x = sqrt(5)/5}}} or {{{x = -sqrt(5)/5}}} 



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Answer:



So the four solutions are:


{{{x = sqrt(15)/3}}}, {{{x = -sqrt(15)/3}}}, {{{x = sqrt(5)/5}}} or {{{x = -sqrt(5)/5}}}