Question 29360
1.sqrt(x-2) = 4
Squaring both the sides
(x-2) =4^2    (using [sqrt(p)]^2 = p and here p = (x-2) )
x-2 = 16
x = 16+2
x=18
Answer: x= 18
Verification:sqrt(x-2) =sqrt(18-2)= sqrt(16) = 4 which is what is given.

2.   2[sqrt(x+3)] = 10
Squaring both the sides,
4(x+3) =10^2     (using [sqrt(p)]^2 = p and here p = (x+3) )
4x+12 = 100
4x= 100-12
4x=88
x=88/4=22
Answer: x=22
Verification:
2[sqrt(x+3)] =22[sqrt(22+3)]=2[sqrt(25)] =2X(5) = 10 which is correct 


3. x = sqrt(x+6)
Squaring both the sides,
x^2 = (x+6)     (using [sqrt(p)]^2 = p and here p = (x+6) )
x^2-x-6=0 ----(*)Which is a quadratic in x
x^2-3x+2x-6 =0 
(x^2-3x)+(2x-6) =0 (by additive associativity)
x(x-3)+2(x-3)
(x-3)(x+2)= 0
(x-3) =0 implies x=3 and 
(x+2)= 0 implies x= -2
Answer: x=3 and x=-2
Verification: We may verify orally and see that both the values hold.
Note:(On the LHS of (*) splitting the middle term two terms 
whose sum is the mid term and whose product is the product of the 
square term and the constant term. 
So here (-x) = (-3x+2x) and (-3x)X(2x) = -6x^2 = (x^2)X(-6)

4. sqrt(x) + 1 = 5
sqrt(x) = 5 - 1
sqrt(x) =4
Squaring both the sides
x= 16     (using [sqrt(p)]^2 = p and here p = (x) )
Answer: x=16
Verification: Very clear orally even!