Question 102534
{{{x^2=-5x-5}}} Start with the given quadratic



{{{0=-x^2-5x-5}}} Subtract {{{x^2}}} from both sides



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^2-5*x-5=0}}} ( notice {{{a=-1}}}, {{{b=-5}}}, and {{{c=-5}}})





{{{x = (--5 +- sqrt( (-5)^2-4*-1*-5 ))/(2*-1)}}} Plug in a=-1, b=-5, and c=-5




{{{x = (5 +- sqrt( (-5)^2-4*-1*-5 ))/(2*-1)}}} Negate -5 to get 5




{{{x = (5 +- sqrt( 25-4*-1*-5 ))/(2*-1)}}} Square -5 to get 25  (note: remember when you square -5, you must square the negative as well. This is because {{{(-5)^2=-5*-5=25}}}.)




{{{x = (5 +- sqrt( 25+-20 ))/(2*-1)}}} Multiply {{{-4*-5*-1}}} to get {{{-20}}}




{{{x = (5 +- sqrt( 5 ))/(2*-1)}}} Combine like terms in the radicand (everything under the square root)




{{{x = (5 +- sqrt(5))/(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 = (5 +- sqrt(5))/-2}}} Multiply 2 and -1 to get -2


So now the expression breaks down into two parts


{{{x = (5 + sqrt(5))/-2}}} or {{{x = (5 - sqrt(5))/-2}}}



Now break up the fraction



{{{x=+5/-2+sqrt(5)/-2}}} or {{{x=+5/-2-sqrt(5)/-2}}}



Simplify



{{{x=-5 / 2-sqrt(5)/2}}} or {{{x=-5 / 2+sqrt(5)/2}}}



So these expressions approximate to


{{{x=-3.61803398874989}}} or {{{x=-1.38196601125011}}}



So our solutions are:

{{{x=-3.61803398874989}}} or {{{x=-1.38196601125011}}}


Notice when we graph {{{-x^2-5*x-5}}}, we get:


{{{ graph( 500, 500, -13.6180339887499, 8.61803398874989, -13.6180339887499, 8.61803398874989,-1*x^2+-5*x+-5) }}}


when we use the root finder feature on a calculator, we find that {{{x=-3.61803398874989}}} and {{{x=-1.38196601125011}}}.So this verifies our answer