`(1)/((t-2)^(2))=6+(1)/(t-2)`
If p and q represents the solutions of the equations above, what is the value of `ptimesq`?

Options

A. `-(3)/(2)`
B. `(7)/(2)`
C. `(9)/(4)`
D. `(15)/(6)`

Answer

To solve the equation \(\frac{1}{(t-2)^2} = 6 + \frac{1}{t-2}\), we will follow these steps:

### Step 1: Eliminate the fractions
Multiply both sides of the equation by \((t-2)^2\) to eliminate the fractions:

\[ 1 = 6(t-2)^2 + (t-2) \]

### Step 2: Expand the right side
Now, expand the right side of the equation:

\[ 1 = 6(t^2 - 4t + 4) + (t - 2) \]
\[ 1 = 6t^2 - 24t + 24 + t - 2 \]
\[ 1 = 6t^2 - 23t + 22 \]

### Step 3: Rearrange the equation
Rearranging the equation gives us:

\[ 0 = 6t^2 - 23t + 21 \]

### Step 4: Identify coefficients
Now, we can identify the coefficients \(a\), \(b\), and \(c\) from the standard form of a quadratic equation \(ax^2 + bx + c = 0\):
- \(a = 6\)
- \(b = -23\)
- \(c = 21\)

### Step 5: Calculate the product of the roots
The product of the roots \(p\) and \(q\) of the quadratic equation can be found using the formula:

\[ p \cdot q = \frac{c}{a} \]

Substituting the values of \(c\) and \(a\):

\[ p \cdot q = \frac{21}{6} = \frac{7}{2} \]

### Final Answer
Thus, the value of \(p \cdot q\) is \(\frac{7}{2}\).

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