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