To solve the equation \( 2x - 1 = \sqrt{8 - x} \), we will start by squaring both sides to eliminate the square root:
\[ (2x - 1)^2 = (\sqrt{8 - x})^2 \]
This gives us:
\[ 4x^2 - 4x + 1 = 8 - x \]
Now, rearranging this equation to bring all terms to one side:
\[ 4x^2 - 4x + x + 1 - 8 = 0 \]
Simplifying further:
\[ 4x^2 - 3x - 7 = 0 \]
Next, we will use the quadratic formula to solve for \( x \):
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 4 \), \( b = -3 \), and \( c = -7 \).
Calculating the discriminant:
\[ b^2 - 4ac = (-3)^2 - 4(4)(-7) = 9 + 112 = 121 \]
Now, substituting back into the quadratic formula:
\[ x = \frac{3 \pm \sqrt{121}}{2 \times 4} \]
\[ x = \frac{3 \pm 11}{8} \]
Calculating the two possible solutions:
- \( x = \frac{14}{8} = \frac{7}{4} \)
- \( x = \frac{-8}{8} = -1 \)
Hence, the possible solutions are \( x = \frac{7}{4} \) and \( x = -1 \).
Now, we need to check for extraneous solutions by substituting these values back into the original equation.
Checking \( x = \frac{7}{4} \):
\[ 2\left(\frac{7}{4}\right) - 1 \stackrel{?}{=} \sqrt{8 - \frac{7}{4}} \]
Calculating the left side:
\[ 2\left(\frac{7}{4}\right) - 1 = \frac{14}{4} - 1 = \frac{14}{4} - \frac{4}{4} = \frac{10}{4} = \frac{5}{2} \]
Calculating the right side:
\[ 8 - \frac{7}{4} = \frac{32}{4} - \frac{7}{4} = \frac{25}{4} \implies \sqrt{\frac{25}{4}} = \frac{5}{2} \]
Since both sides match, \( x = \frac{7}{4} \) is a valid solution.
Checking \( x = -1 \):
\[ 2(-1) - 1 \stackrel{?}{=} \sqrt{8 - (-1)} \]
Calculating the left side:
\[ 2(-1) - 1 = -2 - 1 = -3 \]
Calculating the right side:
\[ 8 - (-1) = 8 + 1 = 9 \implies \sqrt{9} = 3 \]
Since -3 does not equal 3, \( x = -1 \) is an extraneous solution.
Thus, the only valid solution is:
\[ \boxed{\frac{7}{4}} \]