Question 257532
1. {{{d(x)=f(x)-g(x)=(8+x)-(x-5)=8+x-x+5=13}}}. 


So {{{d(x)=13}}}

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2. 




{{{x^2+6x+6=0}}} Start with the given equation.



Notice that the quadratic {{{x^2+6x+6}}} is in the form of {{{Ax^2+Bx+C}}} where {{{A=1}}}, {{{B=6}}}, and {{{C=6}}}



Let's use the quadratic formula to solve for "x":



{{{x = (-B +- sqrt( B^2-4AC ))/(2A)}}} Start with the quadratic formula



{{{x = (-(6) +- sqrt( (6)^2-4(1)(6) ))/(2(1))}}} Plug in  {{{A=1}}}, {{{B=6}}}, and {{{C=6}}}



{{{x = (-6 +- sqrt( 36-4(1)(6) ))/(2(1))}}} Square {{{6}}} to get {{{36}}}. 



{{{x = (-6 +- sqrt( 36-24 ))/(2(1))}}} Multiply {{{4(1)(6)}}} to get {{{24}}}



{{{x = (-6 +- sqrt( 12 ))/(2(1))}}} Subtract {{{24}}} from {{{36}}} to get {{{12}}}



{{{x = (-6 +- sqrt( 12 ))/(2)}}} Multiply {{{2}}} and {{{1}}} to get {{{2}}}. 



{{{x = (-6 +- 2*sqrt(3))/(2)}}} Simplify the square root  (note: If you need help with simplifying square roots, check out this <a href=http://www.algebra.com/algebra/homework/Radicals/simplifying-square-roots.solver> solver</a>)  



{{{x = (-6)/(2) +- (2*sqrt(3))/(2)}}} Break up the fraction.  



{{{x = -3 +- sqrt(3)}}} Reduce.  



{{{x = -3+sqrt(3)}}} or {{{x = -3-sqrt(3)}}} Break up the expression.  



So the solutions are {{{x = -3+sqrt(3)}}} or {{{x = -3-sqrt(3)}}} 



which approximate to {{{x=-1.268}}} or {{{x=-4.732}}} 


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3. 



Simply plug in {{{x=1/4}}} to get {{{f(1/4) = e^(2(1/4))=e^(2/4)=e^(1/2)^""}}}



So {{{f(1/4) = e^(1/2)^""}}}. In addition, you can convert to radical notation to get   {{{f(1/4) = sqrt(e)}}}