# Solving perfect square quadratic inequality

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!

• $y^2>0\implies y>0$ or $y<0$ Commented Jul 18, 2023 at 14:02
• @J.W.Tanner Thanks! What if I had written it as just (2x-3)(2x-3)>0 though?
– HM88
Commented Jul 18, 2023 at 14:08
• That means $2x-3>0$ or $2x-3<0$ Commented Jul 18, 2023 at 14:19

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):