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: Divide the whole expression by $Q^2$ instead of substituting a value for $P$.
Start from the target expression:
\[ \frac{P^2+Q^2}{P^2-Q^2} \]
Divide every term, top and bottom, by $Q^2$:
\[ \frac{\frac{P^2}{Q^2}+1}{\frac{P^2}{Q^2}-1} \]

Step 2: Rewrite $\frac{P^2}{Q^2}$ as $\left(\frac{P}{Q}\right)^2$.
This is a useful identity since it lets us plug in the given ratio directly:
\[ \frac{P^2}{Q^2} = \left(\frac{P}{Q}\right)^2 \]

Step 3: Substitute the known ratio $\frac{P}{Q} = 8$.
\[ \left(\frac{P}{Q}\right)^2 = 8^2 = 64 \]
So the expression becomes
\[ \frac{64+1}{64-1} \]

Step 4: Simplify.
\[ \frac{64+1}{64-1} = \frac{65}{63} \]

Step 5: Confirm this cannot be reduced further.
Since $65 = 5 \times 13$ and $63 = 9 \times 7$ share no common factor, $\frac{65}{63}$ is already in its simplest form, matching option (c).
\[ \boxed{\dfrac{65}{63}} \]