The equation $k{x}^2-10x+(4k-9)=0$ has roots that are reciprocals of each other. Then $k$ equals:

If the roots are r and 1/r their product is 1, and the product of the roots is $(4k-9)/k$. So $4k-9=k$, giving $3k=9$ and $k=3$. (9/4 comes from setting the constant term $4k-9$ to zero; 5/2 from setting $4k-9$ equal to 1 rather than to $k$; 10/3 from equating the sum of the roots $10/k$ to 3.) At $k=3$ the equation is $3x^2-10x+3=0=(3x-1)(x-3)$, whose roots 3 and 1/3 are reciprocals.