Question

If \(\frac{P}{Q} = 8\), then find the value of \(\frac{P^2+Q^2}{P^2-Q^2}\).

• A. \(\frac{22}{21}\)
• B. \(\frac{9}{7}\)
• C. \(\frac{65}{63}\)
• D. \(\frac{50}{47}\)

Hint:

Substitute \(P = 8Q\) into the expression and cancel \(Q^2\) from top and bottom.

Answer 1

The Correct Option is C

Step 1: Express one variable in terms of the other.
We are given \(\frac{P}{Q} = 8\), so
\[ P = 8Q \]

Step 2: Substitute \(P = 8Q\) into the required expression.
\[ \frac{P^2+Q^2}{P^2-Q^2} = \frac{(8Q)^2+Q^2}{(8Q)^2-Q^2} \]

Step 3: Simplify the squares.
\[ (8Q)^2 = 64Q^2 \]
So the expression becomes
\[ \frac{64Q^2+Q^2}{64Q^2-Q^2} = \frac{65Q^2}{63Q^2} \]

Step 4: Cancel the common factor \(Q^2\).
Since \(Q \ne 0\), the \(Q^2\) in the numerator and denominator cancels:
\[ \frac{65Q^2}{63Q^2} = \frac{65}{63} \]

Step 5: Match with the options.
Option (a) \(\frac{22}{21}\), option (b) \(\frac{9}{7}\) and option (d) \(\frac{50}{47}\) do not equal \(\frac{65}{63}\), so none of them fit. Only option (c) matches exactly.

Final Answer:
The value of the expression is \(\frac{65}{63}\).
\[ \boxed{\dfrac{65}{63}} \]