The equation whose roots are such that their A.M. = 9 and G.M. = 4 is

Options

A. `x^(2)+18x+16=0`
B. `x^(2)-18x+16=0`
C. `x^(2)+18x-16=0`
D. `x^(2)-18x-16=0`

Answer

To find the quadratic equation whose roots have an arithmetic mean (A.M.) of 9 and a geometric mean (G.M.) of 4, we can follow these steps:

### Step 1: Understand the definitions of A.M. and G.M.
The arithmetic mean of two numbers \( \alpha \) and \( \beta \) is given by:
\[ \text{A.M.} = \frac{\alpha + \beta}{2} \]
The geometric mean of two numbers \( \alpha \) and \( \beta \) is given by:
\[ \text{G.M.} = \sqrt{\alpha \beta} \]

### Step 2: Set up equations based on the given means.
Given that the A.M. is 9, we can set up the equation:
\[ \frac{\alpha + \beta}{2} = 9 \]
Multiplying both sides by 2 gives:
\[ \alpha + \beta = 18 \quad \text{(Equation 1)} \]

Given that the G.M. is 4, we can set up the equation:
\[ \sqrt{\alpha \beta} = 4 \]
Squaring both sides gives:
\[ \alpha \beta = 16 \quad \text{(Equation 2)} \]

### Step 3: Form the quadratic equation.
Using the relationships from Equations 1 and 2, we can form the quadratic equation. The standard form of a quadratic equation with roots \( \alpha \) and \( \beta \) is:
\[ x^2 - (\alpha + \beta)x + \alpha \beta = 0 \]
Substituting the values from Equations 1 and 2:
\[ x^2 - 18x + 16 = 0 \]

### Step 4: Write the final quadratic equation.
Thus, the required quadratic equation is:
\[ \boxed{x^2 - 18x + 16 = 0} \]

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