The method of intervals for quadratic inequalities

Question

I want to ask you something about the following quadratic inequality: $$9x^2+12x+4\le 0.$$ Let us find the "=0" points. Here we have $9x^2+12x+4=9(x+\frac{2}{3})^2\ge0$ (we can factor this by calculating the discriminant $D=36-36=0$ and using $ax^{2}+bx+c=a(x-x_1)(x-x_2)$). I am practising the wavy curve method (the method of intervals) to solve quadratic inequalities, so I would want to solve it using this way. What is the general method to approach the quadratic inequality when the discriminant is zero, and we want to use the wavy curve method?

Answer 1

In this case there is no need to find the intersection with $y=0$, because you re considering the inequality: $$9(x+\frac{2}{3})^2\ge0$$ which is always true for every $x$ in $R$. On the other side, when you have an inequality in the form: $$ax^2+bx+c\geq0$$ you can always rewite it as: $$a(x-x_1)(x-x_2)\geq0$$ and then study the sign of $x-x_1$ and $x-x_2$. The interval where both factors are positive or both are negative are correct and are solutions to the inequality.

Also, I recommend you to know this theorem:

Give a polynomial $P(x)=a^+bx+c$, the sign of $P(x)$ betweehn the roots $(P(x)=0)$ is opposite respect to the sign outsider the interval of the roots.