Whose perfect square is \(49u^2+14uv+v^2\)?

Correct answer: \((7u+v)^2\)

The direct answer is A: \((7u+v)^2\). Use the identity \((p+q)^2=p^2+2pq+q^2\). The first term \(49u^2\) is \((7u)^2\), and the last term \(v^2\) is \(v^2\). The middle term predicted by the identity is \(2(7u)(v)=14uv\), exactly the middle term in the question. Therefore \(49u^2+14uv+v^2=(7u+v)^2\). A is correct because its expansion gives all three terms with the required signs. B would begin with \(2401u^2\), not \(49u^2\). C expands to \(49u^2-14uv+v^2\), so its middle sign is wrong. D expands to \(u^2+14uv+49v^2\), so the first and last terms are interchanged. A useful check is that the middle term must be twice the product of the square roots of the first and last square terms.