Finding the Product of Roots (\(\alpha\beta\)) for the Radical Equation

We are asked to find the value of \(\alpha\beta\), where \(\alpha\) and \(\beta\) are the roots of the equation:

\[ \sqrt{3 x^2-7 x-30}-\sqrt{2 x^2-7 x-5} = x - 5 \]

The key steps involve algebraic manipulation, including squaring both sides, to transform the radical equation into a simpler polynomial form, likely a quadratic equation, whose roots can then be analyzed.

Detailed Algebraic Solution

1. Rearrange the Equation: First, isolate one of the square root terms. Let's move the second square root term to the right side of the equation:

   \[ \sqrt{3 x^2-7 x-30} = \sqrt{2 x^2-7 x-5} + (x - 5) \]

2. Square Both Sides: Squaring both sides helps eliminate the primary square root on the left.

   \[ \left(\sqrt{3 x^2-7 x-30}\right)^2 = \left(\sqrt{2 x^2-7 x-5} + (x - 5)\right)^2 \]

   \[ 3 x^2-7 x-30 = \left(\sqrt{2 x^2-7 x-5}\right)^2 + (x - 5)^2 + 2 \left(\sqrt{2 x^2-7 x-5}\right)(x - 5) \]

   Expand the terms:

   \[ 3 x^2-7 x-30 = (2 x^2-7 x-5) + (x^2 - 10x + 25) + 2 (x - 5) \sqrt{2 x^2-7 x-5} \]

   Combine like terms on the right side:

   \[ 3 x^2-7 x-30 = (2x^2 + x^2) + (-7x - 10x) + (-5 + 25) + 2 (x - 5) \sqrt{2 x^2-7 x-5} \]

   \[ 3 x^2-7 x-30 = 3 x^2 - 17x + 20 + 2 (x - 5) \sqrt{2 x^2-7 x-5} \]

3. Isolate the Remaining Square Root Term: Group the non-radical terms on one side.

   \[ (3 x^2-7 x-30) - (3 x^2 - 17x + 20) = 2 (x - 5) \sqrt{2 x^2-7 x-5} \]

   Simplify the left side:

   \[ (3x^2 - 3x^2) + (-7x + 17x) + (-30 - 20) = 2 (x - 5) \sqrt{2 x^2-7 x-5} \]

   \[ 10x - 50 = 2 (x - 5) \sqrt{2 x^2-7 x-5} \]

   Factor the left side:

   \[ 10(x - 5) = 2 (x - 5) \sqrt{2 x^2-7 x-5} \]

4. Solve for x by Cases:
   • Case 1: If \(x - 5 = 0\), which means \(x = 5\). We should verify this potential root in the original equation.
   • Case 2: If \(x - 5 \neq 0\), we can divide both sides by \(2(x - 5)\).

     \[ \frac{10(x - 5)}{2(x - 5)} = \sqrt{2 x^2-7 x-5} \]

     \[ 5 = \sqrt{2 x^2-7 x-5} \]

5. Square Again (Case 2): Square both sides of the equation \(5 = \sqrt{2 x^2-7 x-5}\) to eliminate the remaining square root.

   \[ 5^2 = \left(\sqrt{2 x^2-7 x-5}\right)^2 \]

   \[ 25 = 2 x^2-7 x-5 \]

   Rearrange this into a standard quadratic form (\(ax^2+bx+c=0\)):

   \[ 2 x^2-7 x-5 - 25 = 0 \]

   \[ 2 x^2-7 x-30 = 0 \]

6. Calculate the Product of Roots (\(\alpha\beta\)): The quadratic equation \(2 x^2-7 x-30 = 0\) potentially contains the roots of the original equation. Using Vieta's formulas, the product of the roots (\(\alpha\beta\)) of a quadratic equation \(ax^2 + bx + c = 0\) is given by the ratio \(c/a\).

   For the equation \(2 x^2-7 x-30 = 0\), we have \(a=2\), \(b=-7\), and \(c=-30\).

   Therefore, the product of the roots is:

   \[ \alpha\beta = \frac{c}{a} = \frac{-30}{2} \]

   \[ \alpha\beta = -15 \]

Final Result Interpretation

Following the algebraic steps, we arrived at the quadratic equation \(2 x^2-7 x-30 = 0\). The product of its roots is \(-15\). This value corresponds to option 1.