Question 169983


{{{w^2-5w+5=0}}} Start with the given equation.



Notice we have a quadratic equation in the form of {{{aw^2+bw+c}}} where {{{a=1}}}, {{{b=-5}}}, and {{{c=5}}}



Let's use the quadratic formula to solve for w



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



{{{w = (-(-5) +- sqrt( (-5)^2-4(1)(5) ))/(2(1))}}} Plug in  {{{a=1}}}, {{{b=-5}}}, and {{{c=5}}}



{{{w = (5 +- sqrt( (-5)^2-4(1)(5) ))/(2(1))}}} Negate {{{-5}}} to get {{{5}}}. 



{{{w = (5 +- sqrt( 25-4(1)(5) ))/(2(1))}}} Square {{{-5}}} to get {{{25}}}. 



{{{w = (5 +- sqrt( 25-20 ))/(2(1))}}} Multiply {{{4(1)(5)}}} to get {{{20}}}



{{{w = (5 +- sqrt( 5 ))/(2(1))}}} Subtract {{{20}}} from {{{25}}} to get {{{5}}}



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



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



So the answers are {{{w = (5+sqrt(5))/(2)}}} or {{{w = (5-sqrt(5))/(2)}}} 



which approximate to {{{w=3.618}}} or {{{w=1.382}}}