Question

Sqrt( 7x ) + 1 = sqrt( 7x + 1 )

Solve it?

Answer 1

`√(7x) +1 = √(7x+1)`

First isolate a square root on the left side:

`√(7x) = - 1 + √(7x+1)`

Now, delete the radical on the left side, Raise both sides squarely:

`(√(7x))^2 = (-1+ √(7x+1) )^2`


Thus:

`7x = 1 - 2 √(7x+1) + 7x + 1`

`7x = 7x+1+1-2 √(7x+1)`

`7x = 7x+2-2 √(7x+1)`

Find the radical remainder by isolating a radical on the left side again:

`2 √(7x+1) = -\diagup\!\!\!\! 7x+\diagup\!\!\!\! 7x+2`

`2 √(7x+1) = 2`

Now, delete the radical on the left side, Raise both sides squarely:

`(2 √(7x+1))^2 = (2)^2`

Solving, We have:

`2^2(7x+1) = 4`

`4(7x+1) =4`

`28x+4 = 4`

`28x = \diagup\!\!\!\! 4 -\diagup\!\!\!\! 4`

`28x = 0`

`x = (0)/(28)`

`\boxed{x = 0}`

Confirm that the solution is correct
Statement equation

`√(7x) = - 1 + √(7x+1)`

*Replaces "0" in "x"

`√(7*0) = - 1 + √(7*0+1)`

`√(0) = - 1 + √(0+1)`

`0 = - 1 + √(1)`

`0 = - \diagup\!\!\!\! 1 + \diagup\!\!\!\! 1`
Solution:

`\boxed{0 = 0}` (TRUE)