Question 236187


{{{x^2-x=1}}} Start with the given equation.



{{{x^2-x-1=0}}} Subtract 1 from both sides.



Notice that the quadratic {{{x^2-x-1}}} is in the form of {{{Ax^2+Bx+C}}} where {{{A=1}}}, {{{B=-1}}}, and {{{C=-1}}}



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



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



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



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



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



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



{{{x = (1 +- sqrt( 1+4 ))/(2(1))}}} Rewrite {{{sqrt(1--4)}}} as {{{sqrt(1+4)}}}



{{{x = (1 +- sqrt( 5 ))/(2(1))}}} Add {{{1}}} to {{{4}}} to get {{{5}}}



{{{x = (1 +- sqrt( 5 ))/(2)}}} Multiply {{{2}}} and {{{1}}} to get {{{2}}}. 



{{{x = (1+sqrt(5))/(2)}}} or {{{x = (1-sqrt(5))/(2)}}} Break up the expression.  



So the solutions are {{{x = (1+sqrt(5))/(2)}}} or {{{x = (1-sqrt(5))/(2)}}} 



which approximate to {{{x=1.618}}} or {{{x=-0.618}}}