Solving Quadratic Inequalities with 2 or 1 Roots

Suppose we have a quadratic function $f(x)$ with $1$ or $2$ real roots. In this scenario, we can use special methods to solve an inequality like $f(x) \geq 0$.

Let's focus on the inequality $f(x) \geq 0$ for now. We can address $f(x) \leq 0$ later since it is an identical process and intuition.

Example

Consider $f(x) = x^2-4x+3$. From Factoring Polynomials, we know we can write $f(x) = (x-3)(x-1)$. So, we know $f(x)$ has roots at $x=1$ and $x=3$.

Now, suppose we want to find the $x$ values such that $f(x) \geq 0$. We can first build intuition by looking at the graph for $f(x)$

Now, just looking at the graph, we can see that for $x \leq 1$ and $x \geq 3$, $f(x) \geq 0$.

If we want to do this algebraically, we will take advantage of the fact that $f(x)$ has two real roots, and can be written as $f(x) = (x-3)(x-1)$.

Notice that for $(x-3)(x-1)$ to be non-negative, either both $x-3$ and $x-1$ have to be non-positive or both $x-3$ and $x-1$ have to be non-negative. We can write this as $$\begin{align*} \begin{cases} x-3 \leq 0\\ x-1\leq 0 \end{cases} \text{ or } \begin{cases} x-3 \geq 0\\ x-1\geq 0 \end{cases} \end{align*}$$

Now, we just have to solve each of these systems of linear inequalities separately, and we can get a solution for $f(x) \geq 0$ by combining the solutions.

For the first one, we see $x-3 \leq 0$ holds for $x \leq 3$ and $x-1 \leq 0$ holds for $x \leq 1$. Therefore, the solution to the entire system of inequalities is when both hold, which is when $x \leq 1$.

For the second one, we see $x-3 \geq 0$ holds for $x \geq 3$ and $x-1 \geq 0$ holds for $x \geq 1$. Therefore, the solution to the entire system of inequalities is when both hold, which is when $x \geq 3$.

Since we only need one of these to hold to make $f(x) \geq 0$, we know that $f(x) \geq 0$ for $x \leq 1$ and $x \geq 3$

Other inequalities

Now, if we want to extend this to solving $f(x) \leq 0$, we just have to notice that $f(x) \leq 0$ precisely when $$\begin{align*} \begin{cases} x-3 \geq 0\\ x-1\leq 0 \end{cases} \text{ or } \begin{cases} x-3 \leq 0\\ x-1\geq 0 \end{cases} \end{align*}$$

Then, we can solve this system like we did above and get a solution for $f(x) \leq 0$.

Only One Root

But what if there is only $1$ root? Then, we can use the same process as above, where we factor and solve the systems of linear inequalities that arise. This is because we know any quadratic function which has exactly $1$ root can be written in the form $f(x) = a(x-c)^2$, where $c$ is the root.

Summary

If you have a function $f(x)$ which has $1$ or $2$ real roots, do the following steps to try and solve an inequality of the form $f(x) \geq 0$, $f(x) \leq 0$, $f(x) > 0$ or $f(x) < 0$.

1. Factor $f(x)$ as $f(x) = a(x-c)(x-d)$.

2. Depending on the inequality you're solving, convert the inequality to systems of linear inequalities in terms of $x-c$ and $x-d$.

3. The solutions to the systems of linear inequalities you get should be the same as the solutions to the original inequality, so you're all done!