There are two quadratic equations `E_1` and `E_2` Each root of `E_1` is four times to that of `E_2`, and the minimum of `E_1` will be m times to-that of `E_2` Then m is

Options

A. 0
B. 2
C. 16
D. Cannot be determined uniquely

Answer

To solve the problem, we need to derive the relationship between the minimum values of the two quadratic equations \(E_1\) and \(E_2\) based on the given conditions. Here’s a step-by-step solution:

### Step 1: Define the Quadratic Equations
Let the roots of \(E_2\) be \(\alpha\) and \(\beta\). Then, the roots of \(E_1\) will be \(4\alpha\) and \(4\beta\) since each root of \(E_1\) is four times that of \(E_2\).

The quadratic equations can be expressed as:
- \(E_2: x^2 - (\alpha + \beta)x + \alpha\beta = 0\)
- \(E_1: x^2 - (4\alpha + 4\beta)x + (4\alpha)(4\beta) = 0\)

### Step 2: Calculate the Coefficients
From the roots, we can identify the coefficients:
- For \(E_2\):
- \(a_2 = 1\)
- \(b_2 = -(\alpha + \beta)\)
- \(c_2 = \alpha\beta\)

- For \(E_1\):
- \(a_1 = 1\)
- \(b_1 = -4(\alpha + \beta)\)
- \(c_1 = 16\alpha\beta\)

### Step 3: Find the Minimum Values
The minimum value of a quadratic equation \(ax^2 + bx + c\) occurs at \(x = -\frac{b}{2a}\) and is given by:
\[ \text{Minimum} = -\frac{b^2}{4a} + c \]

#### Minimum of \(E_2\):
\[ \text{Minimum of } E_2 = -\frac{(-(\alpha + \beta))^2}{4 \cdot 1} + \alpha\beta = -\frac{(\alpha + \beta)^2}{4} + \alpha\beta \]

#### Minimum of \(E_1\):
\[ \text{Minimum of } E_1 = -\frac{(-4(\alpha + \beta))^2}{4 \cdot 1} + 16\alpha\beta = -\frac{16(\alpha + \beta)^2}{4} + 16\alpha\beta = -4(\alpha + \beta)^2 + 16\alpha\beta \]

### Step 4: Express the Minimum of \(E_1\) in Terms of \(E_2\)
Now, we need to relate the minimum of \(E_1\) to that of \(E_2\):
\[ \text{Minimum of } E_1 = -4(\alpha + \beta)^2 + 16\alpha\beta \]
\[ \text{Minimum of } E_2 = -\frac{(\alpha + \beta)^2}{4} + \alpha\beta \]

### Step 5: Find the Ratio \(m\)
To find \(m\), we express the minimum of \(E_1\) in terms of the minimum of \(E_2\):
\[ \text{Minimum of } E_1 = 16 \left(-\frac{(\alpha + \beta)^2}{4} + \alpha\beta\right) = 16 \cdot \text{Minimum of } E_2 \]
Thus, we have:
\[ m = 16 \]

### Final Answer
The value of \(m\) is \(16\).