Question

matk311_mathematics iii semester a_2026-080220207-integ math 3a polynomials, rational, and radical relationships: part three what is the extraneous solution to the equation \\(\sqrt{2p + 1} + 2\sqrt{p} = 1\\)? \\(p = -4\\) \\(p = -2\\) \\(p = 4\\) \\(p = 0\\) 01:53:20

Explanation:

Step1: Analyze Domain Restrictions

For the equation \(\sqrt{2p + 1}+2\sqrt{p}=1\), the radicands must be non - negative. So we have the following inequalities:
- For \(\sqrt{2p + 1}\), \(2p+1\geq0\), which gives \(p\geq-\frac{1}{2}\).
- For \(\sqrt{p}\), \(p\geq0\).
Combining these two, the domain of the equation is \(p\geq0\).

Step2: Check Each Solution

- **For \(p = - 4\)**:
Since \(-4<0\), it is not in the domain of the original equation. But we still check it in the transformed equation (after squaring). However, from the domain analysis, we know it is extraneous because it does not satisfy the domain requirements of the square - root functions in the original equation.
- **For \(p=-2\)**:
\(-2 < 0\), so it is not in the domain of the original equation.
- **For \(p = 4\)**:
Substitute \(p = 4\) into the original equation:
Left - hand side (LHS)=\(\sqrt{2\times4 + 1}+2\sqrt{4}=\sqrt{9}+2\times2=3 + 4=7\)
Right - hand side (RHS) = 1. Since \(7 eq1\), we will further analyze by solving the equation properly. But first, let's transform the equation.
Start with \(\sqrt{2p + 1}+2\sqrt{p}=1\). Rearrange it as \(\sqrt{2p + 1}=1 - 2\sqrt{p}\).
Square both sides: \(2p + 1=(1 - 2\sqrt{p})^2=1-4\sqrt{p}+4p\).
Simplify: \(2p + 1=1-4\sqrt{p}+4p\)
Subtract \(2p + 1\) from both sides: \(0 = 2p-4\sqrt{p}\)
Factor: \(0 = 2\sqrt{p}(\sqrt{p}-2)\)
So \(\sqrt{p}=0\) or \(\sqrt{p}=2\), which gives \(p = 0\) or \(p = 4\).
Now check \(p = 4\) in the original equation: \(\sqrt{2\times4 + 1}+2\sqrt{4}=\sqrt{9}+4 = 3 + 4=7 eq1\). So \(p = 4\) is extraneous.
- **For \(p = 0\)**:
Substitute \(p = 0\) into the original equation:
LHS=\(\sqrt{2\times0 + 1}+2\sqrt{0}=\sqrt{1}+0 = 1\)
RHS = 1. So \(p = 0\) is a valid solution.

Among the given options, \(p = 4\) is an extraneous solution. We can also see that \(p=-4\) and \(p = - 2\) are extraneous due to domain issues, but \(p = 4\) is a solution to the squared equation but not to the original equation. Since the question asks for the extraneous solution from the given options, and considering the process of solving the equation (after squaring we get \(p = 0\) and \(p = 4\), and \(p = 4\) does not satisfy the original equation), and also the domain - based extraneous solutions (\(p=-4\) and \(p=-2\)) are outside the domain. But when we solve the equation by squaring, the non - domain solutions are also extraneous. However, the most relevant extraneous solution from the process of solving (after considering the domain and the squaring process) is \(p = 4\). But wait, let's re - check the domain. The domain is \(p\geq0\), so \(p=-4\) and \(p=-2\) are out of the domain. But when we solve the equation by squaring, we get \(p = 0\) and \(p = 4\). \(p = 0\) is valid, \(p = 4\) is not. Also, \(p=-4\) and \(p=-2\) are extraneous because they are not in the domain. But the question is about the extraneous solution to the equation. Let's check the original equation's domain again. The term \(\sqrt{p}\) requires \(p\geq0\), so \(p=-4\) and \(p=-2\) are extraneous because they don't satisfy the domain of \(\sqrt{p}\). But when we solve the equation by moving terms and squaring, we get \(p = 0\) and \(p = 4\). \(p = 4\) is extraneous as it doesn't satisfy the original equation. But among the options, \(p = 4\) is the one that comes from the squaring process and is extraneous. Wait, let's substitute \(p = 4\) into the original equation: \(\sqrt{2*4 + 1}+2\sqrt{4}=\sqrt{9}+4 = 3 + 4=7 eq1\). Substitute \(p = 0\): \(\sqrt{1}+0 = 1\), which works. Now, for \(p=-4\): \(\sqrt{2*(-4)+1}+2\sqrt{-4}\), but \(\sqrt{-4}\) is not a real number, so it's extraneous. But in the context of the equation (assuming real - valued solutions), the extraneous solutions from the squaring process (after considering the domain) are \(p = 4\) (since \(p=-4\) and \(p=-2\) are not in the domain of \(\sqrt{p}\)). So the extraneous solution from the given options (after proper solving) is \(p = 4\). Wait, but let's check the options again. The options are \(p=-4\), \(p=-2\), \(p = 4\), \(p = 0\). The domain is \(p\geq0\), so \(p=-4\) and \(p=-2\) are extraneous due to domain. But when we solve the equation, we get \(p = 0\) (valid) and \(p = 4\) (extraneous). So the extraneous solution is \(p = 4\).

Answer:

\(p = 4\) (the option with \(p = 4\))