Question 90108
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^2-7*x-2=0}}} ( notice {{{a=1}}}, {{{b=-7}}}, and {{{c=-2}}})


{{{x = (--7 +- sqrt( (-7)^2-4*1*-2 ))/(2*1)}}} Plug in a=1, b=-7, and c=-2




{{{x = (7 +- sqrt( (-7)^2-4*1*-2 ))/(2*1)}}} Negate -7 to get 7




{{{x = (7 +- sqrt( 49-4*1*-2 ))/(2*1)}}} Square -7 to get 49  (note: remember when you square -7, you must square the negative as well. This is because {{{(-7)^2=-7*-7=49}}}.)




{{{x = (7 +- sqrt( 49+8 ))/(2*1)}}} Multiply {{{-4*-2*1}}} to get {{{8}}}




{{{x = (7 +- sqrt( 57 ))/(2*1)}}} Combine like terms in the radicand (everything under the square root)




{{{x = (7 +- sqrt(57))/(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 = (7 +- sqrt(57))/2}}} Multiply 2 and 1 to get 2


So now the expression breaks down into two parts


{{{x = (7 + sqrt(57))/2}}} or {{{x = (7 - sqrt(57))/2}}}



Now break up the fraction



{{{x=+7/2+sqrt(57)/2}}} or {{{x=+7/2-sqrt(57)/2}}}



Simplify



{{{x=7 / 2+sqrt(57)/2}}} or {{{x=7 / 2-sqrt(57)/2}}}



So these expressions approximate to


{{{x=7.27491721763537}}} or {{{x=-0.274917217635375}}}



So our solutions are:

{{{x=7.27491721763537}}} or {{{x=-0.274917217635375}}}


Notice when we graph {{{x^2-7*x-2}}}, we get:


{{{ graph( 500, 500, -10.2749172176354, 17.2749172176354, -10.2749172176354, 17.2749172176354,1*x^2+-7*x+-2) }}}


when we use the root finder feature on a calculator, we find that {{{x=7.27491721763537}}} and {{{x=-0.274917217635375}}}.So this verifies our answer