Question 90125
Starting with the general quadratic


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


the general solution using the quadratic equation is:


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


So lets solve {{{x^2-4*x-8=0}}} ( notice {{{a=1}}}, {{{b=-4}}}, and {{{c=-8}}})


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




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




{{{x = (4 +- sqrt( 16-4*1*-8 ))/(2*1)}}} Square -4 to get 16  (note: remember when you square -4, you must square the negative as well. This is because {{{(-4)^2=-4*-4=16}}}.)




{{{x = (4 +- sqrt( 16+32 ))/(2*1)}}} Multiply {{{-4*-8*1}}} to get {{{32}}}




{{{x = (4 +- sqrt( 48 ))/(2*1)}}} Combine like terms in the radicand (everything under the square root)




{{{x = (4 +- 4*sqrt(3))/(2*1)}}} Simplify the square root (note: If you need help with simplifying the square root, check out this <a href=http://www.algebra.com/algebra/homework/Radicals/simplifying-square-roots.solver> solver</a>)




{{{x = (4 +- 4*sqrt(3))/2}}} Multiply 2 and 1 to get 2


So now the expression breaks down into two parts


{{{x = (4 + 4*sqrt(3))/2}}} or {{{x = (4 - 4*sqrt(3))/2}}}



Now break up the fraction



{{{x=+4/2+4*sqrt(3)/2}}} or {{{x=+4/2-4*sqrt(3)/2}}}



Simplify



{{{x=2+2*sqrt(3)}}} or {{{x=2-2*sqrt(3)}}}



So these expressions approximate to


{{{x=5.46410161513775}}} or {{{x=-1.46410161513775}}}



So our solutions are:

{{{x=5.46410161513775}}} or {{{x=-1.46410161513775}}}


Notice when we graph {{{x^2-4*x-8}}}, we get:


{{{ graph( 500, 500, -11.4641016151378, 15.4641016151378, -11.4641016151378, 15.4641016151378,1*x^2+-4*x+-8) }}}


when we use the root finder feature on a calculator, we find that {{{x=5.46410161513775}}} and {{{x=-1.46410161513775}}}.So this verifies our answer