I was taking a test and came across the problem:
solve the following quadratic inequality and express the solution in interval notation$$4x^2 > 12x - 9$$
I got this far:
$$4x^2 - 12x + 9 > 0 $$
$$(2x-3)^2>0$$
$$x > \frac32$$
Then I drew a number line and tested either side of $\frac 32 $ with 5 and -1, and got:
$$(-\infty,\infty)$$
My teacher wrote the answer as:
$$ \left(- {\infty}, \frac {-3}{2} \right)\cup \left( \frac32,\infty \right) $$
So I obviously did something wrong, but my question is, what? And how did she get $\frac {-3}{2}$?
Thanks!
Recall that the square of a negative real number is positive. This means that the inequality $$ (2x-3)^2>0 $$ implies $$ (2x-3)>0\quad\text{or}\quad(2x-3)<0 $$ which holds when $x>\frac32$ or $x<\frac32$. In other words, the original inequality holds for all $x\in\mathbb R$ except at the point $x=\frac32$. Thus, the answer should be $$ \left(- {\infty}, \frac {3}{2} \right)\cup \left( \frac32,\infty \right). $$
A quick sketch of the plot of $y=4x^2-12x+9$ makes the answer intuitive (the entire parabola lies above the $x$-axis except at $x=\frac32$, where it touches the $x$-axis):