Question 201639


{{{8r^2=9}}} Start with the given equation.



{{{8r^2-9=0}}} Subtract 9 from both sides.



Notice that the quadratic {{{8r^2-9}}} is in the form of {{{Ar^2+Br+C}}} where {{{A=8}}}, {{{B=0}}}, and {{{C=-9}}}



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



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



{{{r = (-(0) +- sqrt( (0)^2-4(8)(-9) ))/(2(8))}}} Plug in  {{{A=8}}}, {{{B=0}}}, and {{{C=-9}}}



{{{r = (0 +- sqrt( 0-4(8)(-9) ))/(2(8))}}} Square {{{0}}} to get {{{0}}}. 



{{{r = (0 +- sqrt( 0--288 ))/(2(8))}}} Multiply {{{4(8)(-9)}}} to get {{{-288}}}



{{{r = (0 +- sqrt( 0+288 ))/(2(8))}}} Rewrite {{{sqrt(0--288)}}} as {{{sqrt(0+288)}}}



{{{r = (0 +- sqrt( 288 ))/(2(8))}}} Add {{{0}}} to {{{288}}} to get {{{288}}}



{{{r = (0 +- sqrt( 288 ))/(16)}}} Multiply {{{2}}} and {{{8}}} to get {{{16}}}. 



{{{r = (0 +- 12*sqrt(2))/(16)}}} 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>)  



{{{r = (12*sqrt(2))/(16)}}} or {{{r = -(12*sqrt(2))/(16)}}} Break up the expression.  



{{{r = (3/4)*sqrt(2)}}} or {{{r = -(3/4)*sqrt(2)}}} Reduce.



So the solutions are {{{r = (3/4)*sqrt(2)}}} or {{{r = -(3/4)*sqrt(2)}}}