Question

Solve the radical​ equation, and check all proposed solutions.
StartRoot 19 x minus 46 EndRoot equals x plus 2
Question content area bottom
Part 1
A.
​{5​, 10​}
B.
​{10​}
C.
​{5​}
D.empty set
empty set

Answer 1

To solve the equation \(\sqrt{19x - 46} = x + 2\), we first need to square both sides to eliminate the square root. However, remember that squaring both sides may introduce extraneous solutions, so we will check all potential solutions at the end.

### Step 1: Square both sides

\[
(\sqrt{19x - 46})^2 = (x + 2)^2
\]

This simplifies to:

\[
19x - 46 = (x + 2)(x + 2)
\]

Expanding the right-hand side:

\[
19x - 46 = x^2 + 4x + 4
\]

### Step 2: Rearranging the equation

Rearranging the equation to set it to 0 gives us:

\[
0 = x^2 + 4x + 4 - 19x + 46
\]

Combining like terms:

\[
0 = x^2 - 15x + 50
\]

### Step 3: Solve the quadratic equation

Now, we can solve the quadratic equation \(x^2 - 15x + 50 = 0\). We can use the quadratic formula:

\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

where \(a = 1\), \(b = -15\), and \(c = 50\).

Calculating the discriminant:

\[
b^2 - 4ac = (-15)^2 - 4 \cdot 1 \cdot 50 = 225 - 200 = 25
\]

Now plug it into the quadratic formula:

\[
x = \frac{15 \pm \sqrt{25}}{2(1)} = \frac{15 \pm 5}{2}
\]

Calculating the two possible values for \(x\):

1. \(x = \frac{15 + 5}{2} = \frac{20}{2} = 10\)
2. \(x = \frac{15 - 5}{2} = \frac{10}{2} = 5\)

So the proposed solutions are \(x = 10\) and \(x = 5\).

### Step 4: Check each solution in the original equation

1. **For \(x = 10\)**:

\[
\sqrt{19(10) - 46} = 10 + 2
\]
\[
\sqrt{190 - 46} = 12
\]
\[
\sqrt{144} = 12 \quad \text{(True)}
\]

2. **For \(x = 5\)**:

\[
\sqrt{19(5) - 46} = 5 + 2
\]
\[
\sqrt{95 - 46} = 7
\]
\[
\sqrt{49} = 7 \quad \text{(True)}
\]

### Conclusion

Both proposed solutions \(x = 10\) and \(x = 5\) are valid. Therefore, the solution set is:

\(\{5, 10\}\)

Thus, the answer is:

**A. \(\{5, 10\}\)**