The harmonic mean of two numbers is `-(8)/(5)` and their geometric mean is 2 . The quadratic equation whose roots are twice those numbers is

Options

A. `x^(2)+10x+16=0`
B. `x^(2)-10x+16=0`
C. `x^(2)+5x+4=0`
D. `x^(2)-5x+4=0`

Answer

To find the quadratic equation whose roots are twice the two numbers given their harmonic mean and geometric mean, we can follow these steps:

### Step 1: Define the Variables
Let the two numbers be \( A \) and \( B \).

### Step 2: Use the Harmonic Mean Formula
The harmonic mean (HM) of two numbers \( A \) and \( B \) is given by:
\[ HM = \frac{2AB}{A + B} \]
We know from the problem that:
\[ HM = -\frac{8}{5} \]
Thus, we can set up the equation:
\[ \frac{2AB}{A + B} = -\frac{8}{5} \]

### Step 3: Use the Geometric Mean Formula
The geometric mean (GM) of two numbers \( A \) and \( B \) is given by:
\[ GM = \sqrt{AB} \]
We know from the problem that:
\[ GM = 2 \]
Squaring both sides gives us:
\[ AB = 2^2 = 4 \]

### Step 4: Substitute \( AB \) into the Harmonic Mean Equation
Substituting \( AB = 4 \) into the harmonic mean equation:
\[ \frac{2 \cdot 4}{A + B} = -\frac{8}{5} \]
This simplifies to:
\[ \frac{8}{A + B} = -\frac{8}{5} \]

### Step 5: Solve for \( A + B \)
Cross-multiplying gives:
\[ 8 \cdot 5 = -8(A + B) \]
This simplifies to:
\[ 40 = -8(A + B) \]
Dividing both sides by -8:
\[ A + B = -\frac{40}{8} = -5 \]

### Step 6: Form the Quadratic Equation
We need to find the quadratic equation whose roots are \( 2A \) and \( 2B \). The sum of the roots \( 2A + 2B \) is:
\[ 2(A + B) = 2(-5) = -10 \]
The product of the roots \( 2A \cdot 2B \) is:
\[ 4AB = 4 \cdot 4 = 16 \]

### Step 7: Write the Quadratic Equation
The quadratic equation can be written in the form:
\[ x^2 - (\text{sum of roots})x + (\text{product of roots}) = 0 \]
Substituting the values we found:
\[ x^2 - (-10)x + 16 = 0 \]
This simplifies to:
\[ x^2 + 10x + 16 = 0 \]

### Final Answer
The quadratic equation whose roots are twice the two numbers is:
\[ \boxed{x^2 + 10x + 16 = 0} \]