Suppose that the quadratic equations $3x^{2}+px+1=0$ and $2x^{2}+qx+1=0$ have a common root then the value of $5pa-2p^{2}-3q^{2}$ equals

The problem likely contains a typographical error in the requested expression, which is usually given as the symmetric formula 5pq - 2p^2 - 3q^2. Let r be the common root of the equations 3x^2 + px + 1 = 0 and 2x^2 + qx + 1 = 0, meaning 3r^2 + pr + 1 = 0 and 2r^2 + qr + 1 = 0. Eliminating the constant term 1 by cross-multiplying gives (3r^2 + pr) = (2r^2 + qr), which simplifies to r( r + p - q ) = 0. Since the constant term is 1, r cannot be 0, so we have r = q - p. Substituting r = q - p into the second equation yields 2(q - p)^2 + q(q - p) + 1 = 0, which rearranges to the quadratic equation 3p^2 - 5pq + 3q^2 + 1 = 0. Substituting p = 3 and q = 2 into the symmetric expression 5pq - 2p^2 - 3q^2 gives a result of 0.

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