Question

what is the extraneous solution to the equation \\(\sqrt{2p + 1} + 2\sqrt{p} = 1\\)?
\\(\bigcirc\\ p = -4\\)
\\(\bigcirc\\ p = -2\\)
\\(\bigcirc\\ p = 0\\)
\\(\bigcirc\\ p = 4\\)

Explanation:

Step1: Analyze Domain

For square roots, \(2p + 1\geq0\) and \(p\geq0\). So \(p\geq0\). Eliminate \(p = - 4\), \(p=-2\) (since they are negative).

Step2: Isolate a Radical

Start with \(\sqrt{2p + 1}+2\sqrt{p}=1\). Isolate \(\sqrt{2p + 1}\): \(\sqrt{2p + 1}=1 - 2\sqrt{p}\).

Step3: Square Both Sides

Square both sides: \(2p + 1=(1 - 2\sqrt{p})^2\). Expand RHS: \(2p + 1 = 1-4\sqrt{p}+4p\).

Step4: Simplify Equation

Subtract \(2p + 1\) from both sides: \(0 = 1-4\sqrt{p}+4p-(2p + 1)\). Simplify: \(0 = 2p-4\sqrt{p}\). Factor: \(0 = 2\sqrt{p}(\sqrt{p}-2)\).

Step5: Solve for p

Set factors to zero: \(2\sqrt{p}=0\) (gives \(p = 0\)) or \(\sqrt{p}-2 = 0\) (gives \(p = 4\)).

Step6: Check Solutions

• For \(p = 0\): LHS \(=\sqrt{0 + 1}+2\sqrt{0}=1+0 = 1\), RHS \(=1\). Valid.
• For \(p = 4\): LHS \(=\sqrt{8 + 1}+2\sqrt{4}=3 + 4 = 7\), RHS \(=1\). Not valid. So \(p = 4\) is extraneous.

Answer:

D. \(p = 4\) (assuming the options are labeled as A. \(p=-4\), B. \(p = - 2\), C. \(p = 0\), D. \(p = 4\))