Question

cphs : advanced algebra: concepts and connections - block (27.0831030)
radical equations and extraneous roots
which equation can be rewritten as ( x + 4 = x^2 )? assume ( x > 0 )
( sqrt{x} + 2 = x )
( sqrt{x + 4} = x )
( sqrt{x + 2} = x )
( sqrt{x^2 + 16} = x )

Explanation:

Step1: Analyze each option

For a radical equation \(\sqrt{y}=x\), we can square both sides to get \(y = x^2\) (since \(x>0\), we don't have to consider negative roots for now). We need to find which equation, when squared, gives \(x + 4=x^2\).

Option 1: \(\sqrt{x + 2}=x\)

Square both sides: \((\sqrt{x + 2})^2=x^2\) → \(x + 2=x^2\). Not the desired equation.

Option 2: \(\sqrt{x + 4}=x\)

Square both sides: \((\sqrt{x + 4})^2=x^2\) → \(x + 4=x^2\). This matches the desired equation.

Option 3: \(\sqrt{x^2 + 2}=x\)

Square both sides: \((\sqrt{x^2+2})^2=x^2\) → \(x^2 + 2=x^2\) → \(2 = 0\) (which is false, and also not the desired equation).

Option 4: \(\sqrt{x^2 + 16}=x\)

Square both sides: \((\sqrt{x^2 + 16})^2=x^2\) → \(x^2+16=x^2\) → \(16 = 0\) (false, not desired).

Answer:

\(\boldsymbol{\sqrt{x + 4}=x}\) (the second option in the list of equations)