To solve the equation
[tex]\[
\sqrt{6 x - 38} = \sqrt{2 x - 9} - \sqrt{4 x - 25}
\][/tex]
we need to follow these steps:
1. Isolate one of the square roots:
Given the equation already has isolated square roots on each side, we start manipulating directly.
2. Square both sides:
Squaring both sides helps to remove the square roots:
[tex]\[
(\sqrt{6 x - 38})^2 = \left(\sqrt{2 x - 9} - \sqrt{4 x - 25}\right)^2
\][/tex]
Simplifying this gives:
[tex]\[
6 x - 38 = (\sqrt{2 x - 9})^2 + (\sqrt{4 x - 25})^2 - 2 \sqrt{(2 x - 9)(4 x - 25)}
\][/tex]
Which simplifies to:
[tex]\[
6 x - 38 = (2 x - 9) + (4 x - 25) - 2 \sqrt{(2 x - 9)(4 x - 25)}
\][/tex]
Combining like terms:
[tex]\[
6 x - 38 = 6 x - 34 - 2 \sqrt{(2 x - 9)(4 x - 25)}
\][/tex]
3. Isolate the remaining square root term:
We need to isolate the term containing the square root:
[tex]\[
6 x - 38 - 6 x + 34 = - 2 \sqrt{(2 x - 9)(4 x - 25)}
\][/tex]
Simplifying further:
[tex]\[
-4 = - 2 \sqrt{(2 x - 9)(4 x - 25)}
\][/tex]
Dividing by -2 on both sides:
[tex]\[
2 = \sqrt{(2 x - 9)(4 x - 25)}
\][/tex]
4. Square both sides again:
Squaring both sides to remove the square root gives:
[tex]\[
4 = (2 x - 9)(4 x - 25)
\][/tex]
Expanding and simplifying:
[tex]\[
4 = 8 x^2 - 50 x - 36 x + 225
\][/tex]
Combining like terms:
[tex]\[
4 = 8 x^2 - 86 x + 225
\][/tex]
Bringing all terms to one side to set the equation to zero:
[tex]\[
8 x^2 - 86 x + 221 = 0
\][/tex]
5. Solve the quadratic equation:
Using the quadratic formula [tex]\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex], where [tex]\(a = 8\)[/tex], [tex]\(b = -86\)[/tex], and [tex]\(c = 221\)[/tex]:
[tex]\[
x = \frac{86 \pm \sqrt{(-86)^2 - 4 \cdot 8 \cdot 221}}{2 \cdot 8}
\][/tex]
Simplifying inside the square root and the rest of the formula, we find:
[tex]\[
x = \frac{86 \pm \sqrt{7396 - 7072}}{16}
\][/tex]
[tex]\[
x = \frac{86 \pm \sqrt{324}}{16}
\][/tex]
[tex]\[
x = \frac{86 \pm 18}{16}
\][/tex]
So, we get:
[tex]\[
x = \frac{104}{16} = 6.5 \quad \text{or} \quad x = \frac{68}{16} = 4.25
\][/tex]
6. Verify solutions:
We substitute back into the original equation to verify if [tex]\(x = 6.5\)[/tex] is valid.
Since
[tex]\[
\sqrt{6(6.5) - 38} = \sqrt{2(6.5) - 9} - \sqrt{4(6.5) - 25}
\][/tex]
giving
[tex]\[
\sqrt{39 - 38} = \sqrt{13 - 9} - \sqrt{26-25}
\][/tex]
which simplifies to:
[tex]\[
\sqrt{1} = \sqrt{4} - \sqrt{1}
\][/tex]
[tex]\[
1 = 2 - 1 = 1
\][/tex]
Thus, we confirm [tex]\(x = 6.5\)[/tex] is a solution.
So, the solution set is [tex]\(\left\{\frac{13}{2}\right\}\)[/tex], which corresponds to option B.
Therefore, the correct answer is:
B. The solution set is the empty set.
