Question

Sqrt( 7x ) + 1 = sqrt( 7x + 1 ) Solve it?

Answer 1

$\sqrt{7x} +1 = \sqrt{7x+1}$

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)