# How do you solve the inequality 16x^2+9<24x?

Dec 7, 2016

There are no solutions. (The solution set is empty, $\emptyset$)
$16 {x}^{2} + 9 < 24 x$ if and only if $16 {x}^{2} - 24 x + 9 < 0$
But $16 {x}^{2} - 24 x + 9 = {\left(4 x + 3\right)}^{2}$ which is never negative. (Is never less trhan $0$.)