Question

Solve the following radical equation: 2x−1=8−x−−−−√ Check for extraneous solutions. (1 point) Responses x=74 x is equal to 7 fourths x=-1 x=-1 x=−74 and x=1 x=−74 and x=1 x=−1 and x=74

Answer 1

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:

1. \( x = \frac{14}{8} = \frac{7}{4} \)
2. \( x = \frac{-8}{8} = -1 \)

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:

\[
\boxed{\frac{7}{4}}
\]