Official Solution:If \(f(x) = 2x - 1\) and \(g(x) = x^2\), then what is the product of all values of \(n\) for which \(f(n^2)=g(n+12)\)?A. \(-145\)
B. \(-24\)
C. \(24\)
D. \(145\)
E. \(None \ of \ the \ above\)
\(f(x) = 2x - 1\), hence \(f(n^2)=2n^2-1\).
\(g(x) = x^2\), hence \(g(n+12)=(n+12)^2=n^2+24n+144\).
Since we are given that \(f(n^2)=g(n+12)\), we have the equation \(2n^2-1=n^2+24n+144\). Simplifying this equation gives us \(n^2-24n-145=0\).
Next,
using Vieta's theorem, we know that for the roots \(x_1\) and \(x_2\) of a quadratic equation \(ax^2+bx+c=0\): \(x_1+x_2=\frac{-b}{a}\) and \(x_1*x_2=\frac{c}{a}\).
Applying this theorem to our equation, we get \(n_1*n_2=-145\)
Answer: A