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.