Question 124660
Let's use the quadratic formula to solve for x:



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^2-6*x-3=0}}} ( notice {{{a=1}}}, {{{b=-6}}}, and {{{c=-3}}})





{{{x = (--6 +- sqrt( (-6)^2-4*1*-3 ))/(2*1)}}} Plug in a=1, b=-6, and c=-3




{{{x = (6 +- sqrt( (-6)^2-4*1*-3 ))/(2*1)}}} Negate -6 to get 6




{{{x = (6 +- sqrt( 36-4*1*-3 ))/(2*1)}}} Square -6 to get 36  (note: remember when you square -6, you must square the negative as well. This is because {{{(-6)^2=-6*-6=36}}}.)




{{{x = (6 +- sqrt( 36+12 ))/(2*1)}}} Multiply {{{-4*-3*1}}} to get {{{12}}}




{{{x = (6 +- sqrt( 48 ))/(2*1)}}} Combine like terms in the radicand (everything under the square root)




{{{x = (6 +- 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 = (6 +- 4*sqrt(3))/2}}} Multiply 2 and 1 to get 2


So now the expression breaks down into two parts


{{{x = (6 + 4*sqrt(3))/2}}} or {{{x = (6 - 4*sqrt(3))/2}}}



Now break up the fraction



{{{x=+6/2+4*sqrt(3)/2}}} or {{{x=+6/2-4*sqrt(3)/2}}}



Simplify



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



So these expressions approximate to


{{{x=6.46410161513775}}} or {{{x=-0.464101615137754}}}



So our solutions are:

{{{x=6.46410161513775}}} or {{{x=-0.464101615137754}}}


Notice when we graph {{{x^2-6*x-3}}}, we get:


{{{ graph( 500, 500, -10.4641016151378, 16.4641016151378, -10.4641016151378, 16.4641016151378,1*x^2+-6*x+-3) }}}


when we use the root finder feature on a calculator, we find that {{{x=6.46410161513775}}} and {{{x=-0.464101615137754}}}.So this verifies our answer