find the sum of all real roots to the equation (x-1)^(1/2) + (13-x)^(1/4) = 2[(3)^(1/2)] ans:17, plz explain?

√(x - 1) + ∜(13 - x) = 2√3
Let √(x - 1) = a and ∜(13 - x) = b , then
a + b = 2√3 .... ①
a² + b⁴= 12 ... ②
Substitute ① into ②:
(2√3 - b)² + b⁴= 12
b⁴+ b² - 4√3b + 12 = 12
b(b³ + b - 4√3) = 0
b(b - √3)(b² + √3b + 4) = 0
b₁= 0 , b₂= √3 ,
b₃+ b₄= - √3 (rejected since complex ∜(13 - x) = ki (k > 0) for real x and hence b₃+ b₄is not real)
∜(13 - x) = 0 or ∜(13 - x) = √3
x = 13 or x = 4
The sum of all real roots = 13 + 4 = 17.

In order that the terms of the equation remain real, we have (x-1)>=0 and (13-x)>=0, or
0<=x<=13.
A graph of the function in this interval reveals zeroes at x=4 and x=13.
Check:
when x=4,
(4-1)^(1/2)+(13-4)^(1/4)=sqrt(3)+sqrt(3)=2sqrt(3)
when x=13
(13-1)^(1/2)+(13-13)^(1/4)=sqrt(12)+0=2sqrt(3).

Hence the sum of the real roots is 4+13=17