Solution: Calculating \(\frac{p}{q}+\frac{q}{p}\) for Given Radical Expressions
This problem involves simplifying algebraic expressions containing square roots and performing arithmetic operations.
Understanding the Expressions for \(p\) and \(q\)
We are given the following two expressions:
- \(p=\frac{\sqrt{5}+2}{\sqrt{5}-2}\)
- \(q=\frac{\sqrt{5}-2}{\sqrt{5}+2}\)
The task is to determine the value of the expression \(\frac{p}{q}+\frac{q}{p}\).
Simplifying \(p\) and \(q\) using Rationalization
To simplify the calculations, we first rationalize the denominators of \(p\) and \(q\). Rationalization helps eliminate square roots from the denominator.
Rationalizing the denominator for \(p\)
To rationalize the denominator of \(p\), we multiply both the numerator and the denominator by the conjugate of \(\sqrt{5}-2\), which is \(\sqrt{5}+2\):
\(p = \frac{\sqrt{5}+2}{\sqrt{5}-2} \times \frac{\sqrt{5}+2}{\sqrt{5}+2}\)
We apply the algebraic identities \((a+b)^2 = a^2+b^2+2ab\) for the numerator and \((a-b)(a+b) = a^2-b^2\) for the denominator:
\(p = \frac{(\sqrt{5}+2)^2}{(\sqrt{5})^2 - 2^2} = \frac{(\sqrt{5})^2 + 2^2 + 2(\sqrt{5})(2)}{5 - 4}\)
\(p = \frac{5 + 4 + 4\sqrt{5}}{1} = 9 + 4\sqrt{5}\)
Rationalizing the denominator for \(q\)
Similarly, for \(q\), we multiply the numerator and denominator by the conjugate of \(\sqrt{5}+2\), which is \(\sqrt{5}-2\):
\(q = \frac{\sqrt{5}-2}{\sqrt{5}+2} \times \frac{\sqrt{5}-2}{\sqrt{5}-2}\)
Applying the same identities:
\(q = \frac{(\sqrt{5}-2)^2}{(\sqrt{5})^2 - 2^2} = \frac{(\sqrt{5})^2 + 2^2 - 2(\sqrt{5})(2)}{5 - 4}\)
\(q = \frac{5 + 4 - 4\sqrt{5}}{1} = 9 - 4\sqrt{5}\)
After rationalization, we have \(p = 9 + 4\sqrt{5}\) and \(q = 9 - 4\sqrt{5}\).
Calculating \(\frac{p}{q}+\frac{q}{p}\)
There are a couple of efficient ways to calculate the required expression \(\frac{p}{q}+\frac{q}{p}\).
Method 1: Recognizing \(q\) as the reciprocal of \(p\)
Let's compute the product \(pq\) using the original forms:
\(pq = \left(\frac{\sqrt{5}+2}{\sqrt{5}-2}\right) \times \left(\frac{\sqrt{5}-2}{\sqrt{5}+2}\right) = 1\)
Since \(pq=1\), it follows that \(q = \frac{1}{p}\). Substituting this into the expression \(\frac{p}{q}+\frac{q}{p}\) gives:
\(\frac{p}{q}+\frac{q}{p} = \frac{p}{1/p} + \frac{1/p}{p} = p^2 + \frac{1}{p^2}\)
Because \(q = \frac{1}{p}\), we have \(q^2 = \left(\frac{1}{p}\right)^2 = \frac{1}{p^2}\). Thus, the expression simplifies to \(p^2 + q^2\).
Now, we calculate \(p^2\) and \(q^2\) using their simplified forms:
- \(p^2 = (9 + 4\sqrt{5})^2 = 9^2 + (4\sqrt{5})^2 + 2(9)(4\sqrt{5}) = 81 + (16 \times 5) + 72\sqrt{5} = 81 + 80 + 72\sqrt{5} = 161 + 72\sqrt{5}\)
- \(q^2 = (9 - 4\sqrt{5})^2 = 9^2 + (4\sqrt{5})^2 - 2(9)(4\sqrt{5}) = 81 + 80 - 72\sqrt{5} = 161 - 72\sqrt{5}\)
Adding \(p^2\) and \(q^2\):
\(p^2 + q^2 = (161 + 72\sqrt{5}) + (161 - 72\sqrt{5}) = 161 + 161 = 322\)
Method 2: Using a Common Denominator
Alternatively, we can combine the fractions \(\frac{p}{q}+\frac{q}{p}\) over a common denominator, which is \(pq\):
\(\frac{p}{q}+\frac{q}{p} = \frac{p \cdot p}{q \cdot p} + \frac{q \cdot q}{p \cdot q} = \frac{p^2}{pq} + \frac{q^2}{pq} = \frac{p^2+q^2}{pq}\)
We already established that \(pq=1\). To find \(p^2+q^2\), we can use the algebraic identity \((p+q)^2 = p^2+q^2+2pq\). Rearranging this identity, we get \(p^2+q^2 = (p+q)^2 - 2pq\).
First, let's calculate the sum \(p+q\) using the simplified forms:
\(p+q = (9 + 4\sqrt{5}) + (9 - 4\sqrt{5}) = 9 + 9 = 18\)
Now, substitute the values of \(p+q\) and \(pq\) into the formula for \(p^2+q^2\):
\(p^2+q^2 = (18)^2 - 2(1)\)
\(p^2+q^2 = 324 - 2 = 322\)
Finally, we find the value of the original expression:
\(\frac{p^2+q^2}{pq} = \frac{322}{1} = 322\)
Final Result
Both methods demonstrate that the value of \(\frac{p}{q}+\frac{q}{p}\) is 322.
Similar Questions
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