Verify that \((\sqrt{11} + \sqrt{15})^2\) is an irrational number, given that \(\sqrt{165}\) is an irrational number.
Proof:
\((\sqrt{11} + \sqrt{15})^2\) \(=\) \(11 + 2 \sqrt{165} + 15\)
\(=\) \(26 + 2 \sqrt{165}\)
Let us assume that \(26 + 2 \sqrt{165}\) is a number.
\(26 + 2 \sqrt{165} =\) , where \(p\) and \(q\) are and \( q \neq 0\).
\(\sqrt{165} = \)
Here \(p\) and \(q\) are rationals. So, is a number.
This implies that \(\sqrt{165}\) is a number.
This contradicts the given that \(\sqrt{165}\) is number.
Hence, \((\sqrt{11} + \sqrt{15})^2\) is number.
Answer variants:
\(\frac{q}{p}\)
whole number
\(\frac{p}{q}\)
\(\frac{p-26q}{2q}\)
rational
\(\frac{2p}{q-26p}\)
irrational
\(\frac{q-26p}{2q}\)
integers