To solve the equation [tex]\(2x - 1 = \sqrt{8 - x}\)[/tex] and determine any extraneous solutions, we follow these steps:
1. Square both sides of the equation to get rid of the square root:
[tex]\[
(2x - 1)^2 = (\sqrt{8 - x})^2
\][/tex]
This simplifies to:
[tex]\[
(2x - 1)^2 = 8 - x
\][/tex]
2. Expand the left side of the equation:
[tex]\[
4x^2 - 4x + 1 = 8 - x
\][/tex]
3. Rearrange the equation to bring all terms to one side, forming a standard quadratic equation:
[tex]\[
4x^2 - 4x + 1 - 8 + x = 0
\][/tex]
Simplify:
[tex]\[
4x^2 - 3x - 7 = 0
\][/tex]
4. Solve the quadratic equation [tex]\(4x^2 - 3x - 7 = 0\)[/tex]:
The solutions to this quadratic can be found using the quadratic formula [tex]\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex], where [tex]\(a = 4\)[/tex], [tex]\(b = -3\)[/tex], and [tex]\(c = -7\)[/tex].
5. Substitute these values into the quadratic formula:
[tex]\[
x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4 \cdot 4 \cdot (-7)}}{2 \cdot 4}
\][/tex]
Simplify inside the square root:
[tex]\[
x = \frac{3 \pm \sqrt{9 + 112}}{8}
\][/tex]
[tex]\[
x = \frac{3 \pm \sqrt{121}}{8}
\][/tex]
[tex]\[
x = \frac{3 \pm 11}{8}
\][/tex]
6. Separate the solutions:
[tex]\[
x = \frac{3 + 11}{8} = \frac{14}{8} = \frac{7}{4}
\][/tex]
and
[tex]\[
x = \frac{3 - 11}{8} = \frac{-8}{8} = -1
\][/tex]
7. Check each solution in the original equation [tex]\(2x - 1 = \sqrt{8 - x}\)[/tex] to determine if it is extraneous.
- For [tex]\(x = \frac{7}{4}\)[/tex]:
[tex]\[
2 \left( \frac{7}{4} \right) - 1 = \sqrt{8 - \frac{7}{4}}
\][/tex]
Simplify both sides:
[tex]\[
\frac{14}{4} - 1 = \sqrt{8 - \frac{7}{4}} \implies \frac{14}{4} - 1 = \sqrt{\frac{32 - 7}{4}} \implies \frac{10}{4} = \sqrt{\frac{25}{4}} \implies \frac{5}{2} = \frac{5}{2}
\][/tex]
Hence, [tex]\(x = \frac{7}{4}\)[/tex] is a valid solution.
- For [tex]\(x = -1\)[/tex]:
[tex]\[
2(-1) - 1 = \sqrt{8 - (-1)}
\][/tex]
Simplify both sides:
[tex]\[
-2 - 1 = \sqrt{8 + 1} \implies -3 = \sqrt{9} \implies -3 \neq 3
\][/tex]
Hence, [tex]\(x = -1\)[/tex] is not valid.
So, the extraneous solution obtained by Addison is:
[tex]\[
\boxed{-1}
\][/tex]
