## College Algebra (6th Edition)

Solution set: the empty set, $\emptyset$.
Follow the "Procedure for Solving Polynomial lnequalities",\ p.412: 1. Express the inequality in the form $f(x)<0, f(x)>0, f(x)\leq 0$, or $f(x)\geq 0,$ where $f$ is a polynomial function. $x^{2}-6x+9 < 0$ We recognize the LHS as a square of a difference... $(x-3)^{2} < 0$ ... We do not need to follow the steps any further, since this inequality asks: "For which real numbers is a square negative?" The answer: "For none (for no x)." Solution set: the empty set, $\emptyset$.