x = 6[(x-2)^(1/2)-1]. To find ind x.

Solution :

(x-2)^(1/2) is real only when x-2 > 0 or x >2...........(1)

6[(x-2)^2 +1} = x . So x is an integer part of the left. Therefore,

x < 6{(x-2)^(1/2) -1} < x+ f < x+1 where 0 < f < 1.

x < 6(x-2)^(1/2) - 6 < x+1 .....(2) Or taking the right,

6(x-2)^(1/2) < x+7. Squaring,

36(x-2) > x^2 +14x+49

x^2+(14-36)x+49+72 > 0. or

x^2-22x+121 > 0. Or

(x-11)^2 < 0. Or except x = 11 , for all x the inequality holds.............(3)

From the left of inequality at (2), x < 6{(x-2)^(1/2)-1}. Or

x+6 <6{(x-2)^(1/2)- 1}.

(x+6) < 6(x-2)^(1/2). Squaring both sides,

(x+6)^2 < 36(x-2). Or

x^2+12x+36 -36x+72 < 0.

x^2-24x+108 < 0.

(x-18)(x-6) < 0 .

So x should lie between 6 and 18.. Therefore x>2, x>6, x not equal to 11 and x < 18 . Or

6 <x <18 but xis not equal to 11.

We'll open the bracket and we'll get:

6sqrt(x-2)=x+6

We'll square raise the expression above:

36(x-2)=x^2+12x+36

36x-72=x^2+12x+36

x^2-24x+108=0

We'll apply the quadratic formula:

x1=[24+ sqrt (24^2-4*108)]/2=18

x2=[24- sqrt (24^2-4*108)]/2=6