[tex] \sqrt{7x} +1 = \sqrt{7x+1} [/tex]
First isolate a square root on the left side:
[tex] \sqrt{7x} = - 1 + \sqrt{7x+1} [/tex]
Now, delete the radical on the left side, Raise both sides squarely:
[tex] (\sqrt{7x})^2 = (-1+ \sqrt{7x+1} )^2[/tex]
Thus:
[tex]7x = 1 - 2 \sqrt{7x+1} + 7x + 1[/tex]
[tex]7x = 7x+1+1-2 \sqrt{7x+1} [/tex]
[tex]7x = 7x+2-2 \sqrt{7x+1} [/tex]
Find the radical remainder by isolating a radical on the left side again:
[tex] 2 \sqrt{7x+1} = -\diagup\!\!\!\! 7x+\diagup\!\!\!\! 7x+2 [/tex]
[tex] 2 \sqrt{7x+1} = 2[/tex]
Now, delete the radical on the left side, Raise both sides squarely:
[tex] (2 \sqrt{7x+1})^2 = (2)^2[/tex]
Solving, We have:
[tex]2^2(7x+1) = 4
[/tex]
[tex]4(7x+1) =4
[/tex]
[tex]28x+4 = 4
[/tex]
[tex]28x = \diagup\!\!\!\! 4 -\diagup\!\!\!\! 4
[/tex]
[tex]28x = 0[/tex]
[tex]
x = \frac{0}{28}
[/tex]
[tex]\boxed{x = 0}[/tex]
Confirm that the solution is correct
Statement equation
[tex] \sqrt{7x} = - 1 + \sqrt{7x+1} [/tex]
*Replaces "0" in "x"
[tex]\sqrt{7*0} = - 1 + \sqrt{7*0+1}[/tex]
[tex] \sqrt{0} = - 1 + \sqrt{0+1}
[/tex]
[tex]0 = - 1 + \sqrt{1}[/tex]
[tex]0 = - \diagup\!\!\!\! 1 + \diagup\!\!\!\! 1[/tex]
Solution:
[tex]\boxed{0 = 0}[/tex] (TRUE)
