Question 152870
{{{g(x) = k(x) }}} Start with the given equation.



{{{x^3 + 2x^2= x^3 - x^2 + 6}}} Plug in {{{g(x) = x^3 + 2x^2}}} and {{{k(x) = x^3 - x^2 + 6}}}



{{{0=  - 3x^2 + 6}}} Subtract {{{x^3}}} from both sides. Subtract {{{2x^2}}} from both sides.



Notice we have a quadratic equation in the form of {{{ax^2+bx+c}}} where {{{a=-3}}}, {{{b=0}}}, and {{{c=6}}}



Let's use the quadratic formula to solve for x



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



{{{x = (-(0) +- sqrt( (0)^2-4(-3)(6) ))/(2(-3))}}} Plug in  {{{a=-3}}}, {{{b=0}}}, and {{{c=6}}}



{{{x = (0 +- sqrt( 0-4(-3)(6) ))/(2(-3))}}} Square {{{0}}} to get {{{0}}}. 



{{{x = (0 +- sqrt( 0--72 ))/(2(-3))}}} Multiply {{{4(-3)(6)}}} to get {{{-72}}}



{{{x = (0 +- sqrt( 0+72 ))/(2(-3))}}} Rewrite {{{sqrt(0--72)}}} as {{{sqrt(0+72)}}}



{{{x = (0 +- sqrt( 72 ))/(2(-3))}}} Add {{{0}}} to {{{72}}} to get {{{72}}}



{{{x = (0 +- sqrt( 72 ))/(-6)}}} Multiply {{{2}}} and {{{-3}}} to get {{{-6}}}. 



{{{x = (0 +- 6*sqrt(2))/(-6)}}} Simplify the square root  (note: If you need help with simplifying square roots, check out this <a href=http://www.algebra.com/algebra/homework/Radicals/simplifying-square-roots.solver> solver</a>)  



{{{x = (0)/(-6) +- (6*sqrt(2))/(-6)}}} Break up the fraction.  



{{{x = 0 +- -1*sqrt(2)}}} Reduce.  



{{{x = 0-1*sqrt(2)}}} or {{{x = 0+1*sqrt(2)}}} Break up the expression.  



So the answers are {{{x = 0-1*sqrt(2)}}} or {{{x = 0+1*sqrt(2)}}} 



which approximate to {{{x=-1.414}}} or {{{x=1.414}}} 


So the answer choice is c) x &#8776; ± 1.4