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

neela | Student

Given 4- (x-1)^(1/2) = (x+7)^(1/2). We have to find x.

Solution:

We square both sides :

16 - 2*4(x-1)^(1/2) +x-1 = x+7.

=> 16-1 -7 = = 8(x-1)^(1/2)

=> 8 = 8(x-1)^(1/2). We divide by 8:

=> 1 = (x-1)^1/2).

We  square both sides:

1 = x-1.

1+1 = x.

So x= 2.

giorgiana1976 | Student

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)

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.