If \(p + q = 3\) and \(p - q = 1\), find \(2p^2 + 2q^2 - 4pq\).
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We must evaluate \(2p^2+2q^2-4pq\) given \(p+q=3\) and \(p-q=1\).
Take out the common factor 2: \(2p^2+2q^2-4pq=2(p^2+q^2-2pq)\).
The bracket is a perfect square, since \(p^2-2pq+q^2=(p-q)^2\).
So the expression equals \(2(p-q)^2\).
Substituting the given \(p-q=1\), we get \(2(1)^2=2\times1=2\).
Key concept: recognising the identity \(a^2-2ab+b^2=(a-b)^2\) lets us use \(p-q\) directly, so \(p+q=3\) is not even needed.
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