Answer :
To determine the extraneous solutions to the equation [tex]\(\sqrt{2p + 1} + 2\sqrt{p} = 1\)[/tex], we need to test each of the given values for [tex]\(p\)[/tex] and see if they satisfy the equation. The possible values for [tex]\(p\)[/tex] are [tex]\(-4\)[/tex], [tex]\(-2\)[/tex], [tex]\(0\)[/tex], and [tex]\(4\)[/tex]. Let's evaluate the equation for each of these values:
1. For [tex]\(p = -4\)[/tex]:
- [tex]\(\sqrt{2(-4) + 1} + 2\sqrt{-4}\)[/tex]
- [tex]\(\sqrt{-8 + 1} + 2\sqrt{-4}\)[/tex]
- [tex]\(\sqrt{-7} + 2\sqrt{-4}\)[/tex]
Since [tex]\(\sqrt{-7}\)[/tex] and [tex]\(\sqrt{-4}\)[/tex] involve square roots of negative numbers (which are not real), [tex]\(p = -4\)[/tex] is not a valid solution in the real number domain.
2. For [tex]\(p = -2\)[/tex]:
- [tex]\(\sqrt{2(-2) + 1} + 2\sqrt{-2}\)[/tex]
- [tex]\(\sqrt{-4 + 1} + 2\sqrt{-2}\)[/tex]
- [tex]\(\sqrt{-3} + 2\sqrt{-2}\)[/tex]
Similar to the previous case, [tex]\(\sqrt{-3}\)[/tex] and [tex]\(\sqrt{-2}\)[/tex] are not real numbers, so [tex]\(p = -2\)[/tex] also cannot be a valid solution in the real number domain.
3. For [tex]\(p = 0\)[/tex]:
- [tex]\(\sqrt{2(0) + 1} + 2\sqrt{0}\)[/tex]
- [tex]\(\sqrt{0 + 1} + 2\sqrt{0}\)[/tex]
- [tex]\(\sqrt{1} + 2 \cdot 0\)[/tex]
- [tex]\(1 + 0\)[/tex]
- [tex]\(1\)[/tex]
Here, [tex]\(1 = 1\)[/tex], so [tex]\(p = 0\)[/tex] satisfies the equation.
4. For [tex]\(p = 4\)[/tex]:
- [tex]\(\sqrt{2(4) + 1} + 2\sqrt{4}\)[/tex]
- [tex]\(\sqrt{8 + 1} + 2\sqrt{4}\)[/tex]
- [tex]\(\sqrt{9} + 2 \cdot 2\)[/tex]
- [tex]\(3 + 4\)[/tex]
- [tex]\(7\)[/tex]
Here, [tex]\(7 \neq 1\)[/tex], so [tex]\(p = 4\)[/tex] does not satisfy the equation.
Thus, the values of [tex]\(p\)[/tex] which do not satisfy the equation are the extraneous solutions. These values are:
[tex]\[ p = -4, p = -2, p = 4 \][/tex]
So, the extraneous solutions are [tex]\(-4\)[/tex], [tex]\(-2\)[/tex], and [tex]\(4\)[/tex].
1. For [tex]\(p = -4\)[/tex]:
- [tex]\(\sqrt{2(-4) + 1} + 2\sqrt{-4}\)[/tex]
- [tex]\(\sqrt{-8 + 1} + 2\sqrt{-4}\)[/tex]
- [tex]\(\sqrt{-7} + 2\sqrt{-4}\)[/tex]
Since [tex]\(\sqrt{-7}\)[/tex] and [tex]\(\sqrt{-4}\)[/tex] involve square roots of negative numbers (which are not real), [tex]\(p = -4\)[/tex] is not a valid solution in the real number domain.
2. For [tex]\(p = -2\)[/tex]:
- [tex]\(\sqrt{2(-2) + 1} + 2\sqrt{-2}\)[/tex]
- [tex]\(\sqrt{-4 + 1} + 2\sqrt{-2}\)[/tex]
- [tex]\(\sqrt{-3} + 2\sqrt{-2}\)[/tex]
Similar to the previous case, [tex]\(\sqrt{-3}\)[/tex] and [tex]\(\sqrt{-2}\)[/tex] are not real numbers, so [tex]\(p = -2\)[/tex] also cannot be a valid solution in the real number domain.
3. For [tex]\(p = 0\)[/tex]:
- [tex]\(\sqrt{2(0) + 1} + 2\sqrt{0}\)[/tex]
- [tex]\(\sqrt{0 + 1} + 2\sqrt{0}\)[/tex]
- [tex]\(\sqrt{1} + 2 \cdot 0\)[/tex]
- [tex]\(1 + 0\)[/tex]
- [tex]\(1\)[/tex]
Here, [tex]\(1 = 1\)[/tex], so [tex]\(p = 0\)[/tex] satisfies the equation.
4. For [tex]\(p = 4\)[/tex]:
- [tex]\(\sqrt{2(4) + 1} + 2\sqrt{4}\)[/tex]
- [tex]\(\sqrt{8 + 1} + 2\sqrt{4}\)[/tex]
- [tex]\(\sqrt{9} + 2 \cdot 2\)[/tex]
- [tex]\(3 + 4\)[/tex]
- [tex]\(7\)[/tex]
Here, [tex]\(7 \neq 1\)[/tex], so [tex]\(p = 4\)[/tex] does not satisfy the equation.
Thus, the values of [tex]\(p\)[/tex] which do not satisfy the equation are the extraneous solutions. These values are:
[tex]\[ p = -4, p = -2, p = 4 \][/tex]
So, the extraneous solutions are [tex]\(-4\)[/tex], [tex]\(-2\)[/tex], and [tex]\(4\)[/tex].
