Question 102585
To find the equilibrium price, set the two equations equal to one another


{{{-5p+40=-p^2+30p-8 }}}



{{{cross(-5p+5p+40-40)=-p^2+30p-8 +5p-40}}} Add 5p and subtract 40 from both sides to get everything to one side


{{{0=-p^2+35p-48}}} Combine like terms



Let's use the quadratic formula to solve for p:



Starting with the general quadratic


{{{ap^2+bp+c=0}}}


the general solution using the quadratic equation is:


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




So lets solve {{{-p^2+35*p-48=0}}} ( notice {{{a=-1}}}, {{{b=35}}}, and {{{c=-48}}})





{{{p = (-35 +- sqrt( (35)^2-4*-1*-48 ))/(2*-1)}}} Plug in a=-1, b=35, and c=-48




{{{p = (-35 +- sqrt( 1225-4*-1*-48 ))/(2*-1)}}} Square 35 to get 1225  




{{{p = (-35 +- sqrt( 1225+-192 ))/(2*-1)}}} Multiply {{{-4*-48*-1}}} to get {{{-192}}}




{{{p = (-35 +- sqrt( 1033 ))/(2*-1)}}} Combine like terms in the radicand (everything under the square root)




{{{p = (-35 +- sqrt(1033))/(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>)




{{{p = (-35 +- sqrt(1033))/-2}}} Multiply 2 and -1 to get -2


So now the expression breaks down into two parts


{{{p = (-35 + sqrt(1033))/-2}}} or {{{p = (-35 - sqrt(1033))/-2}}}



Now break up the fraction



{{{p=-35/-2+sqrt(1033)/-2}}} or {{{p=-35/-2-sqrt(1033)/-2}}}



Simplify



{{{p=35 / 2-sqrt(1033)/2}}} or {{{p=35 / 2+sqrt(1033)/2}}}



So these expressions approximate to


{{{p=1.4298413200118}}} or {{{p=33.5701586799882}}}



So our possible solutions are:

{{{p=1.4298413200118}}} or {{{p=33.5701586799882}}}




However, when we plug in {{{p=33.5701586799882}}} into either equation, we get a negative value. So the only solution is {{{p=1.4298413200118}}} which rounds to $1.43. So the answer is C