Question 170212
Let {{{z=x^2 -3}}}. So this means that {{{z^2=(x^2 -3)^2}}}



So the equation {{{(x^2 -3)^2 + (x^2 -3) -2=0}}} simplifies to {{{z^2+z-2=0}}}





{{{z^2+z-2=0}}} Start with the given equation.



Notice we have a quadratic equation in the form of {{{az^2+bz+c}}} where {{{a=1}}}, {{{b=1}}}, and {{{c=-2}}}



Let's use the quadratic formula to solve for z



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



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



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



{{{z = (-1 +- sqrt( 1--8 ))/(2(1))}}} Multiply {{{4(1)(-2)}}} to get {{{-8}}}



{{{z = (-1 +- sqrt( 1+8 ))/(2(1))}}} Rewrite {{{sqrt(1--8)}}} as {{{sqrt(1+8)}}}



{{{z = (-1 +- sqrt( 9 ))/(2(1))}}} Add {{{1}}} to {{{8}}} to get {{{9}}}



{{{z = (-1 +- sqrt( 9 ))/(2)}}} Multiply {{{2}}} and {{{1}}} to get {{{2}}}. 



{{{z = (-1 +- 3)/(2)}}} Take the square root of {{{9}}} to get {{{3}}}. 



{{{z = (-1 + 3)/(2)}}} or {{{z = (-1 - 3)/(2)}}} Break up the expression. 



{{{z = (2)/(2)}}} or {{{z =  (-4)/(2)}}} Combine like terms. 



{{{z = 1}}} or {{{z = -2}}} Simplify. 



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Now remember, we let {{{z=x^2-3}}}. 



{{{z = 1}}} Go back to the first equation



{{{x^2-3 = 1}}} Plug in {{{z=x^2-3}}} 



{{{x^2=1+3}}}Add {{{3}}} to both sides.



{{{x^2=4}}} Combine like terms.



{{{x=0+-sqrt(4)}}} Take the square root of both sides.



{{{x=sqrt(4)}}} or {{{x=-sqrt(4)}}} Break up the "plus/minus" to form two equations.



{{{x=2}}} or {{{x=-2}}}  Take the square root of {{{4}}} to get {{{2}}}.




So the first two solutions are {{{x=2}}} or {{{x=-2}}}.



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{{{z = -2}}} Now move onto the second equation



{{{x^2-3 = -2}}} Plug in {{{z=x^2-3}}}



{{{x^2=-2+3}}}Add {{{3}}} to both sides.



{{{x^2=1}}} Combine like terms.



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



{{{x=sqrt(1)}}} or {{{x=-sqrt(1)}}} Break up the "plus/minus" to form two equations.



{{{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 solutions are 


{{{x=2}}}, {{{x=-2}}}, {{{x=1}}} or {{{x=-1}}} 



Note: the order of the solutions does not matter.