Question 96922
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 {{{2*x^2+11*x-1=0}}} ( notice {{{a=2}}}, {{{b=11}}}, and {{{c=-1}}})


{{{x = (-11 +- sqrt( (11)^2-4*2*-1 ))/(2*2)}}} Plug in a=2, b=11, and c=-1




{{{x = (-11 +- sqrt( 121-4*2*-1 ))/(2*2)}}} Square 11 to get 121  




{{{x = (-11 +- sqrt( 121+8 ))/(2*2)}}} Multiply {{{-4*-1*2}}} to get {{{8}}}




{{{x = (-11 +- sqrt( 129 ))/(2*2)}}} Combine like terms in the radicand (everything under the square root)




{{{x = (-11 +- sqrt(129))/(2*2)}}} 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 = (-11 +- sqrt(129))/4}}} Multiply 2 and 2 to get 4


So now the expression breaks down into two parts


{{{x = (-11 + sqrt(129))/4}}} or {{{x = (-11 - sqrt(129))/4}}}



Now break up the fraction



{{{x=-11/4+sqrt(129)/4}}} or {{{x=-11/4-sqrt(129)/4}}}



Simplify



{{{x=-11 / 4+sqrt(129)/4}}} or {{{x=-11 / 4-sqrt(129)/4}}}



So these expressions approximate to


{{{x=0.0894541729001368}}} or {{{x=-5.58945417290014}}}



So our solutions are:

{{{x=0.0894541729001368}}} or {{{x=-5.58945417290014}}}


Notice when we graph {{{2*x^2+11*x-1}}}, we get:


{{{ graph( 500, 500, -15.5894541729001, 10.0894541729001, -15.5894541729001, 10.0894541729001,2*x^2+11*x+-1) }}}


when we use the root finder feature on a calculator, we find that {{{x=0.0894541729001368}}} and {{{x=-5.58945417290014}}}.So this verifies our answer