Chapter 11 - Quadratic Functions and Equations - 11.3 Studying Solutions of Quadratic Equations - 11.3 Exercise Set - Page 718: 17

There is one repeated rational solution.

$ 9t^{2}-48t+64=0\qquad$.... $a=9,\ b=-48,\ c=64$ $ b^{2}-4ac\qquad$....substitute $b$ for $-48,\ a$ for $9$ and $c$ for $64$ $=(-48)^{2}-4\cdot 9\cdot 64$ $=2304-2304$ $=0$ Since the discriminant equals $0$, there is one repeated rational solution.