Identify extraneous solution. see picture above.

x = 0 .

-20/19 (see below).

Step-by-step explanation:

√(x + 1) + √5x = √(7x + 1)

Squaring both sides:

x + 1 + 5x + 2√(x + 1)√5x = 7x + 1

2√(x + 1)√5x = 7x +1 - 1 - x - 5x = x

Squaring both sides:

4(x + 1)(5x) = x^2

20x^2 + 20x - x^2 = 0

19x^2 + 20x) = 0

x(19x + 20) = 0

x = 0, -20/19.

If we substitute x = 0 into the original equation we get

√(0 + 1) + √5*0 = √(7*0 + 1)

1 = 1  so this is a true solution.

-20/19 :- √(-20/19 + 1) is not real so unless we allow complex square root this is extraneous.

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