Question 88259
{{{x^2 = 5x + 8 }}}


{{{x^2 - 5x - 8=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-8=0}}} (notice {{{a=1}}}, {{{b=-5}}}, and {{{c=-8}}})


{{{x = (--5 +- sqrt( (-5)^2-4*1*-8 ))/(2*1)}}} Plug in a=1, b=-5, and c=-8




{{{x = (5 +- sqrt( (-5)^2-4*1*-8 ))/(2*1)}}} Negate -5 to get 5




{{{x = (5 +- sqrt( 25-4*1*-8 ))/(2*1)}}} Square -5 to get 25




{{{x = (5 +- sqrt( 25+32 ))/(2*1)}}} Multiply {{{-4*-8*1}}} to get {{{32}}}




{{{x = (5 +- sqrt( 57 ))/(2*1)}}} Combine like terms in the radicand (everything under the square root)




{{{x = (5 +- sqrt(57))/(2*1)}}} Simplify the square root




{{{x = (5 +- sqrt(57))/2}}} Multiply 2 and 1 to get 2


So now the expression breaks down into two parts


{{{x = (5 + sqrt(57))/2}}} or {{{x = (5 - sqrt(57))/2}}}



Which approximate to


{{{x=6.27491721763537}}} or {{{x=-1.27491721763537}}}



So our solutions are:

{{{x=6.27491721763537}}} or {{{x=-1.27491721763537}}}


Notice when we graph {{{x^2-5*x-8}}} we get:


{{{ graph( 500, 500, -11.2749172176354, 16.2749172176354, -11.2749172176354, 16.2749172176354,1*x^2+-5*x+-8) }}}


when we use the root finder feature on a calculator, we find that {{{x=6.27491721763537}}} and {{{x=-1.27491721763537}}}.So this verifies our answer.