Equation (1):`2x^(2)+7x=4`
Equation (2): `(y-1)^(2)=9`
If f is the greater of the two roots of Equations (1) and g is the lesser of the two roots of Equation (2), what is the value of the product `ftimesg`?

Options

A. `-4`
B. `-1`
C. `2`
D. `8`

Answer

To solve the problem, we need to find the roots of the two given equations and then calculate the product of the specified roots.

### Step 1: Solve Equation (1)
The first equation is:
\[ 2x^2 + 7x = 4 \]
We can rearrange it to standard form:
\[ 2x^2 + 7x - 4 = 0 \]

### Step 2: Factor the Quadratic Equation
To factor the quadratic equation, we need to find two numbers that multiply to \(2 \times (-4) = -8\) and add up to \(7\). The numbers \(8\) and \(-1\) satisfy this condition.

We can rewrite the middle term:
\[ 2x^2 + 8x - x - 4 = 0 \]
Now, we can group the terms:
\[ (2x^2 + 8x) + (-x - 4) = 0 \]
Factoring out common terms:
\[ 2x(x + 4) - 1(x + 4) = 0 \]
This gives us:
\[ (2x - 1)(x + 4) = 0 \]

### Step 3: Find the Roots of Equation (1)
Setting each factor to zero gives:
1. \(2x - 1 = 0 \Rightarrow x = \frac{1}{2}\)
2. \(x + 4 = 0 \Rightarrow x = -4\)

Thus, the roots of Equation (1) are \(x = \frac{1}{2}\) and \(x = -4\). The greater root \(f\) is:
\[ f = \frac{1}{2} \]

### Step 4: Solve Equation (2)
The second equation is:
\[ (y - 1)^2 = 9 \]
Taking the square root of both sides, we have:
\[ y - 1 = 3 \quad \text{or} \quad y - 1 = -3 \]

### Step 5: Find the Roots of Equation (2)
Solving these gives:
1. \(y - 1 = 3 \Rightarrow y = 4\)
2. \(y - 1 = -3 \Rightarrow y = -2\)

Thus, the roots of Equation (2) are \(y = 4\) and \(y = -2\). The lesser root \(g\) is:
\[ g = -2 \]

### Step 6: Calculate the Product \(f \times g\)
Now, we can find the product of \(f\) and \(g\):
\[ f \times g = \frac{1}{2} \times (-2) = -1 \]

### Final Answer
The value of the product \(f \times g\) is:
\[ \boxed{-1} \]