Question 226867
{{{sqrt(2r^2 - 121)}}} = r
Square both sides and you have
2r^2 - 121 = r^2
2r^2 - r^2 = + 121
r^2 = 121
r = {{{sqrt(121)}}}
r = +11 is the valid solution
;
;
{{{4 + sqrt(m-2)}}} = m
{{{sqrt(m-2)}}} = m - 4
Square both sides
 m - 2 = (m-4)^2
m - 2 = m^2 - 8m + 16
0 = m^2 - 8m - m + 16 + 2
A quadratic equation
m^2 - 9m + 18
Factors to
(m-6)(m-3) = 0
m = 6
and 
m = 3
:
You check both solutions in original equation, only one is valid