Find the product of the real roots of the equation, `x^2+18x +30=2sqrt(x^2+18x+45)`

Options

A. Product of real root of th equation is 10
B. Product of real root of th equation is 20
C. Sum of real roots of the equation is `-18`
D. Sum of real roots of the equation is `18`

Answer

To solve the equation \(x^2 + 18x + 30 = 2\sqrt{x^2 + 18x + 45}\), we will follow these steps:

### Step 1: Rewrite the equation
We start with the equation:
\[ x^2 + 18x + 30 = 2\sqrt{x^2 + 18x + 45} \]

### Step 2: Substitute for simplification
Let \(T = x^2 + 18x + 30\). Then, we can rewrite the equation as:
\[ T = 2\sqrt{T + 15} \]

### Step 3: Square both sides
To eliminate the square root, we square both sides:
\[ T^2 = 4(T + 15) \]

### Step 4: Rearrange the equation
Expanding the right side gives:
\[ T^2 = 4T + 60 \]
Rearranging this, we get:
\[ T^2 - 4T - 60 = 0 \]

### Step 5: Factor the quadratic equation
Next, we can factor the quadratic equation:
\[ T^2 - 10T + 6T - 60 = 0 \]
Grouping terms:
\[ T(T - 10) + 6(T - 10) = 0 \]
Factoring out \((T - 10)\):
\[ (T - 10)(T + 6) = 0 \]

### Step 6: Solve for T
Setting each factor to zero gives:
\[ T - 10 = 0 \quad \text{or} \quad T + 6 = 0 \]
Thus, we find:
\[ T = 10 \quad \text{or} \quad T = -6 \]
Since \(T\) must be non-negative (as it represents a squared term), we discard \(T = -6\) and keep \(T = 10\).

### Step 7: Substitute back for x
Now we substitute back for \(T\):
\[ x^2 + 18x + 30 = 10 \]
This simplifies to:
\[ x^2 + 18x + 20 = 0 \]

### Step 8: Calculate the discriminant
To find the roots, we calculate the discriminant \(D\):
\[ D = b^2 - 4ac = 18^2 - 4 \cdot 1 \cdot 20 = 324 - 80 = 244 \]
Since \(D > 0\), there are two real roots.

### Step 9: Find the product of the roots
The product of the roots of the quadratic equation \(ax^2 + bx + c = 0\) is given by:
\[ \text{Product of roots} = \frac{c}{a} = \frac{20}{1} = 20 \]

### Final Answer
Thus, the product of the real roots of the equation is:
\[ \boxed{20} \]