Question 87585
{{{x^2=5x-1}}}


{{{x^2-5x+1=0}}} Get everything to one side


Now let's use the quadratic formula to solve for x:

Starting with the general quadratic


{{{ax^2+bx+c=0}}}


the general form of the quadratic equation is:


{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a)}}}


So lets solve {{{x^2-5*x+1=0}}}


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




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




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




{{{x = (5 +- sqrt( 21 ))/(2*1)}}} Combine like terms in the radicand (everything under the square root)




{{{x = (5 +- sqrt(21))/(2*1)}}} Simplify the square root




{{{x = (5 +- sqrt(21))/2}}} Multiply 2 and 1 to get 2


So now the expression breaks down into two parts


{{{x = (5 + sqrt(21))/2}}} or {{{x = (5 - sqrt(21))/2}}}



Which approximate to


{{{x=4.79128784747792}}} or {{{x=0.20871215252208}}}



So our solutions are:

{{{x=4.79128784747792}}} or {{{x=0.20871215252208}}}


Notice when we graph {{{x^2-5*x+1}}} we get:


{{{ graph( 500, 500, -9.79128784747792, 14.7912878474779, -9.79128784747792, 14.7912878474779,1*x^2+-5*x+1) }}}


when we use the root finder feature on a calculator, we find that {{{x=4.79128784747792}}} and {{{x=0.20871215252208}}}.So this verifies our answer