Question 86430
Use the quadratic formula to solve for x


Starting with the general quadratic


{{{ax^2+bx+c}}}


the general form of the quadratic equation is:


{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a)}}}


So lets solve {{{2*x^2+10*x+11}}}


{{{x = (-10 +- sqrt( (10)^2-4*2*11 ))/(2*2)}}} Plug in a=2, b=10, and c=11




{{{x = (-10 +- sqrt( 100-4*2*11 ))/(2*2)}}} Square 10 to get 100




{{{x = (-10 +- sqrt( 100+-88 ))/(2*2)}}} Multiply {{{-4*11*2}}} to get {{{-88}}}




{{{x = (-10 +- sqrt( 12 ))/(2*2)}}} Combine like terms in the radicand (everything under the square root)




{{{x = (-10 +- 2*sqrt(3))/(2*2)}}} Simplify the square root




{{{x = (-10 +- 2*sqrt(3))/4}}} Multiply 2 and 2 to get 4


So now the expression breaks down into two parts


{{{x = (-10 + 2*sqrt(3))/4}}} or {{{x = (-10 - 2*sqrt(3))/4}}}



Which approximate to


{{{x=-1.63397459621556}}} or {{{x=-3.36602540378444}}}



So our solutions are:

{{{x=-1.63397459621556}}} or {{{x=-3.36602540378444}}}


Notice when we graph {{{2*x^2+10*x+11}}} we get:


{{{ graph( 500, 500, -13.3660254037844, 8.36602540378444, -13.3660254037844, 8.36602540378444,2*x^2+10*x+11) }}}


when we use the root finder feature on our calculator, we find that {{{x=-1.63397459621556}}} and {{{x=-3.36602540378444}}}.So this verifies our answer