What is x if 4-square root ( x-1 ) = square root ( x+7 ) ?

Conditions of existenceof square
roots:

x - 1 >= 0

x
>= 1

x >= -7

The
interval of admissible values of x: [1 ; +infinite).

We'll
add sqrt(x-1) both sides and we'll use the symmetric
property:

sqrt (x+7) + sqrt(x-1) = 4
(1)

We'll multiply the expression of the left side by its
adjoint:

[sqrt (x+7) + sqrt(x-1)]*[sqrt (x+7) - sqrt(x-1)]
= 4*[sqrt (x+7) - sqrt(x-1)]

We'll get a difference of
squares to the left side:

(x+7) - (x-1) = 4*[sqrt (x+7) -
sqrt(x-1)]

We'll remove the brackets from the left
side

x + 7 - x +1= 4*[sqrt (x+7) -
sqrt(x-1)]

We'll eliminate and combine like
terms:

8 = 4*[sqrt (x+7) -
sqrt(x-1)]

2= [sqrt (x+7) - sqrt(x-1)]
(2)

We'll add (1) + (2):

sqrt
(x+7) + sqrt(x-1) + sqrt (x+7) - sqrt(x-1) = 6

We'll
eliminate and combine like terms:

2sqrt (x+7)=
6

sqrt (x+7) = 3

We'll raise
to square both sides:

x + 7 =
9

x = 9 - 7

x =
2

Since the value of x belongs to the range
[1 ; +infinite), we'll accept as solution of the equation: x =
2.