If one zero of $(k+1)x^2-7x+3$ is the reciprocal of the other, the value of $k$ is:

Reciprocal zeroes $p$ and $\frac{1}{p}$ multiply to 1, so the product of the zeroes must equal 1: $\frac{c}{a}=\frac{3}{k+1}=1$, giving $k+1=3$ and $k=2$. (The polynomial is then $3x^2-7x+3$, whose product of zeroes is $\frac{3}{3}=1$.) Answering 3 stops at $k+1=3$ and reports that instead of $k$; 6 solves $k+1=7$, using the middle coefficient rather than the constant; −2 solves $3=-(k+1)$.