If the one of the roots of the equation $a{x^2} + bx + c = 0$ be reciprocal of one of the roots of ${a_1}{x^2} + {b_1}x + {c_1} = 0$, then ${\left( {a{a_1} - c{c_1}} \right)^2} = \left( {b{c_1} - a{b_1}} \right)\left( {{b_1}c - {a_1}b} \right)$.

Let one root of the first equation be m, so the reciprocal root of the second equation is 1/m, meaning m is a root of the second equation multiplied by x squared (giving c1 x squared + b1 x + a1 = 0). Since the quadratic equations a x squared + b x + c = 0 and c1 x squared + b1 x + a1 = 0 share a common root m, substituting it into both yields a m squared + b m + c = 0 and c1 m squared + b1 m + a1 = 0. Using the cross-multiplication method for common roots, we get m squared divided by (b a1 - c b1) equals m divided by (c c1 - a a1) equals 1 divided by (a b1 - b c1). Replacing m with 1 using the identity that relates these ratios, the square of the denominator (a b1 - b c1) must equal the product of the other two denominators, which expands to (b c1 - a b1)(b1 c - a1 b). Thus, (a a1 - c c1) squared equals (b c1 - a b1)(b1 c - a1 b), proving the statement is true.

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