1 Answer

Nocturno · Answer 1 · 2021-06-21T07:15:02+0000

Answer:

The solutions are x=0 and x=1

Explanation:

Assuming your equation is

√(3x + 1) = x + 1

Square both sides to get:

$3x + 1 = ( {x + 1)}^(2)$

Expand the RHS,

$3x + 1 = {x}^(2) + 2x + 1$

Rewrite in standard quadratic form:

${x}^(2) + 2x - 3x + 1 - 1 = 0$

${x}^(2) - x = 0$

x(x - 1) = 0

x=0 or x=1

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