Question 140238
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+4*x+1=0}}} ( notice {{{a=1}}}, {{{b=4}}}, and {{{c=1}}})





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




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




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




{{{x = (-4 +- sqrt( 12 ))/(2*1)}}} Combine like terms in the radicand (everything under the square root)




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


So now the expression breaks down into two parts


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



Now break up the fraction



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



Simplify



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



So these expressions approximate to


{{{x=-0.267949192431123}}} or {{{x=-3.73205080756888}}}



So our solutions are:

{{{x=-0.267949192431123}}} or {{{x=-3.73205080756888}}}


Notice when we graph {{{x^2+4*x+1}}}, we get:


{{{ graph( 500, 500, -13.7320508075689, 9.73205080756888, -13.7320508075689, 9.73205080756888,1*x^2+4*x+1) }}}


when we use the root finder feature on a calculator, we find that {{{x=-0.267949192431123}}} and {{{x=-3.73205080756888}}}.So this verifies our answer