Question 102616
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 {{{5*x^2+9*x+1=0}}} ( notice {{{a=5}}}, {{{b=9}}}, and {{{c=1}}})





{{{x = (-9 +- sqrt( (9)^2-4*5*1 ))/(2*5)}}} Plug in a=5, b=9, and c=1




{{{x = (-9 +- sqrt( 81-4*5*1 ))/(2*5)}}} Square 9 to get 81  




{{{x = (-9 +- sqrt( 81+-20 ))/(2*5)}}} Multiply {{{-4*1*5}}} to get {{{-20}}}




{{{x = (-9 +- sqrt( 61 ))/(2*5)}}} Combine like terms in the radicand (everything under the square root)





{{{x = (-9 +- sqrt(61))/10}}} Multiply 2 and 5 to get 10


So now the expression breaks down into two parts


{{{x = (-9 + sqrt(61))/10}}} or {{{x = (-9 - sqrt(61))/10}}}



Now break up the fraction



{{{x=-9/10+sqrt(61)/10}}} or {{{x=-9/10-sqrt(61)/10}}}





So these expressions approximate to


{{{x=-0.118975032409335}}} or {{{x=-1.68102496759067}}}



So our solutions are:

{{{x=-0.118975032409335}}} or {{{x=-1.68102496759067}}}


Notice when we graph {{{5*x^2+9*x+1}}}, we get:


{{{ graph( 500, 500, -11.6810249675907, 9.88102496759067, -11.6810249675907, 9.88102496759067,5*x^2+9*x+1) }}}


when we use the root finder feature on a calculator, we find that {{{x=-0.118975032409335}}} and {{{x=-1.68102496759067}}}.So this verifies our answer