Question 100449


{{{2x^2+10x+11=0}}} Start with the given equation



{{{2x^2+10x=-11}}} Subtract 11 from both sides
{{{2(x^2+5x)=-11}}} Factor out the leading coefficient 2.  This step is important since we want the {{{x^2}}} coefficient to be equal to 1.




Take half of the x coefficient 5 to get 2.5 (ie {{{5/2=2.5}}})

Now square 2.5 to get 6.25 (ie {{{(2.5)^2=6.25}}})




{{{2(x^2+5x+6.25)=-11+6.25(2)}}} Add this result (6.25) to the expression {{{x^2+5x}}}  inside the parenthesis. Now the expression {{{x^2+5x+6.25}}}  is a perfect square trinomial. Now add the result (6.25)(2) (remember we factored out a 2) to the right side.




{{{2(x+2.5)^2=-11+6.25(2)}}} Factor {{{x^2+5x+6.25}}} into {{{(x+2.5)^2}}} 



{{{2(x+2.5)^2=-11+12.5}}} Multiply 6.25 and 2 to get 12.5




{{{2(x+2.5)^2=1.5}}} Combine like terms on the right side


{{{(x+2.5)^2=0.75}}} Divide both sides by 2



{{{x+2.5=0+-sqrt(0.75)}}} Take the square root of both sides


{{{x=-2.5+-sqrt(0.75)}}} Subtract 2.5 from both sides to isolate x.


So the expression breaks down to

{{{x=-2.5+sqrt(0.75)}}} or {{{x=-2.5-sqrt(0.75)}}}



So our answer is approximately

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


Here is visual proof


{{{ graph( 500, 500, -10, 10, -10, 10, 2x^2+10x+11) }}} graph of {{{y=2x^2+10x+11}}}



When we use the root finder feature on a calculator, we would find that the x-intercepts are {{{x=-1.63397459621556}}} and {{{x=-3.36602540378444}}}, so this verifies our answer.