We have,
#(x + 5)^(1/2) - (5 - 2x)^(1/4) = 0#
#rArr (x + 5)^(1/2) = (5 - 2x)^(1/4)# [Transpose #(5 - 2x)^(1/4)# to R.H.S]
#rArr ((x + 5)^(1/2))^2 = ((5 - 2x)^(1/4))^2# [Squaring both sides]
#rArr x + 5 = (5 - 2x)^(1/2)#
#rArr (x + 5)^2 = 5 - 2x# [Squaring Again]
#rArr x^2 + 10x + 25 = 5 - 2x# [Expanded the L.H.S]
#rArr x^2 + 10x + 2x + 25 - 5 = cancel5 cancel(- 2x) cancel(+ 2x) cancel(- 5)# [Add #2x - 5# to both sides]
#rArr x^2 + 12x + 20 = 0#
#rArrx^2 + (10 + 2)x + 20 = 0# [Break up #12# as #10 + 2#]
#rArrx^2 + 10x + 2x + 20 = 0#
#rArr x(x + 10) + 2(x + 10) = 0#
#rArr (x + 10)(x + 2) = 0#
So, #x = -10, -2.#
Now, If we substitute the respective values for #x#, we can find the values are extraneous roots or not.
Checking for #-10#:-
L.H.S. :- #(-10 + 5)^(1/2) = sqrt(-5) = 5i#.
And, R.H.S :- #(5 - 2*-10)^(1/4) = (25)^(1/4) = sqrt(5)#
So, L.H.S #!=# R.H.S.
So, #-10# is an extraneous root of this equation.
Checking For #-2#:-
L.H.S. :- #(-2 + 5)^(1/2) = sqrt(3)#.
And, R.H.S :- #(5 - 2*-2)^(1/4) = (9)^(1/4) = sqrt(3)#
So, L.H.S #=# R.H.S.
So, #-2# is not an extraneous root of this equation, but it is a legit root.
Hope This Helps.
