For specific positive numbers m and n, the quadratics 16x^2+36x+56 and (mx+n)^2 differ only in their constant term. What is mn?

NoUReverse Jul 27, 2020

#1**+1 **

Nvm i have answer XD

The expansion of $(mx+n)^2$ is $m^2x^2+2mnx+n^2$. If$$16x^2+36x+56 = m^2x^2+2mnx+n^2,$$we can equate the coefficients of $x^2$ to get $16=m^2$, and thus $m=4$ (we ignore the solution $-4$ because we are given that $m$ is positive). We can also equate the coefficients of $x$ above to get $36=2mn=2(4)n=8n$, and thus $n=\frac{36}{8}=\frac{9}{2}$. So, $mn=4\cdot\frac92 = \boxed{18}$. Solution 2: The expansion of $(mx+n)^2$ is $m^2x^2+2mnx+n^2$. If$$16x^2+36x+56 = m^2x^2+2mnx+n^2,$$we can equate the coefficients of $x$ to get $36=2mn$, and so $mn=\frac{36}{2}=\boxed{18}$.

NoUReverse Jul 27, 2020