Question 202657
{{{2/a=8+4a}}} Start with the given equation.



{{{2=a(8+4a)}}} Multiply both sides by "a".



{{{2=8a+4a^2}}} Distribute



{{{0=8a+4a^2-2}}} Subtract 2 from both sides.



{{{0=4a^2+8a-2}}} Rearrange the terms.



Notice that the quadratic {{{4a^2+8a-2}}} is in the form of {{{Aa^2+Ba+C}}} where {{{A=4}}}, {{{B=8}}}, and {{{C=-2}}}



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



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



{{{a = (-(8) +- sqrt( (8)^2-4(4)(-2) ))/(2(4))}}} Plug in  {{{A=4}}}, {{{B=8}}}, and {{{C=-2}}}



{{{a = (-8 +- sqrt( 64-4(4)(-2) ))/(2(4))}}} Square {{{8}}} to get {{{64}}}. 



{{{a = (-8 +- sqrt( 64--32 ))/(2(4))}}} Multiply {{{4(4)(-2)}}} to get {{{-32}}}



{{{a = (-8 +- sqrt( 64+32 ))/(2(4))}}} Rewrite {{{sqrt(64--32)}}} as {{{sqrt(64+32)}}}



{{{a = (-8 +- sqrt( 96 ))/(2(4))}}} Add {{{64}}} to {{{32}}} to get {{{96}}}



{{{a = (-8 +- sqrt( 96 ))/(8)}}} Multiply {{{2}}} and {{{4}}} to get {{{8}}}. 



{{{a = (-8 +- 4*sqrt(6))/(8)}}} 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>)  



{{{a = (-8+4*sqrt(6))/(8)}}} or {{{a = (-8-4*sqrt(6))/(8)}}} Break up the expression. 



{{{a = (-2+sqrt(6))/(2)}}} or {{{a = (-2+sqrt(6))/(2)}}} Reduce



So the solutions are {{{a = (-2+sqrt(6))/(2)}}} or {{{a = (-2+sqrt(6))/(2)}}} 



which approximate to {{{a=0.225}}} or {{{a=-2.225}}}