Question 196631
{{{(2+-sqrt(212/49))/(4/7)}}} Start with the given expression.



{{{(2+sqrt(212/49))/(4/7)}}} or {{{(2-sqrt(212/49))/(4/7)}}} Break up the "plus/minus" to form two separate expressions.



{{{(2+sqrt(212)/sqrt(49))/(4/7)}}} or {{{(2-sqrt(212)/sqrt(49))/(4/7)}}} Break up the square root.



{{{(2+sqrt(212)/7)/(4/7)}}} or {{{(2-sqrt(212)/7)/(4/7)}}} Take the square root of 49 to get 7



{{{(2+sqrt(4*53)/7)/(4/7)}}} or {{{(2-sqrt(4*53)/7)/(4/7)}}} Factor 212 to get 4*53



{{{(2+(sqrt(4)*sqrt(53))/7)/(4/7)}}} or {{{(2-(sqrt(4)*sqrt(53))/7)/(4/7)}}} Break up the square root.



{{{(2+(2sqrt(53))/7)/(4/7)}}} or {{{(2-(2sqrt(53))/7)/(4/7)}}} Take the square root of 4 to get 2



{{{(2+(2sqrt(53))/7)*(7/4)}}} or {{{(2-(2sqrt(53))/7)*(7/4)}}} Multiply by the reciprocal of the second fraction



{{{2*(7/4)+((2sqrt(53))/7)*(7/4)}}} or {{{2*(7/4)-((2sqrt(53))/7)*(7/4)}}} Distribute




{{{14/4+(14sqrt(53))/(28)}}} or {{{14/4-(14sqrt(53))/(28)}}} Multiply



{{{7/2+sqrt(53)/2}}} or {{{7/2-sqrt(53)/2}}} Reduce



=========================================================================


Answer:



So {{{(2+sqrt(212/49))/(4/7)=7/2+sqrt(53)/2}}} and {{{(2-sqrt(212/49))/(4/7)=7/2-sqrt(53)/2}}}