Defining quadratic inequalities and graphing their intervals

For solving quadratic inequalities we must rember how we can solve quadratic equation. For quadratic equation:

$ ax^2 + bx + c = 0$, the solution is:

$$x_{1,2} = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}$$.

The problem of solving quadratic inequalities is very much connected to solving zeros of quadratic function and determining whether the function is positive or negative.

These are the inequalities that come in form:

$ ax^2 + bx + c > 0$

$ ax^2 + bx + c < 0$.

If variable $x$ satisfies the inequality $ ax^2 + bx + c > 0$ then variable $x$ is located above $x$- axis, and if $x$ satisfies the inequality $ ax^2 + bx + c < 0$ then it is located below $x$ – axis.

And, of course, we know that if $x$ satisfies the equation $ ax^2 + bx + c = 0$ then variable $x$ is the zero of this equation and is located on $x$- axis.

To solve quadratic inequality means to find the set of real numbers $x$ for which function f receives positive or negative numbers depending how our inequality looks.

Example 1. Solve the following inequality.

$ x^2 + x – 2 > 0$

First thing you do when solving quadratic inequalities is the same as when you’re solving quadratic equations. First you find the zeros and draw your function.

$ x^2 + x – 2 = (x + 2)\cdot (x – 1)$

$$x_{1,2} = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}$$

$ x_1 = – 2$ and $ x_2 = 1$

The leading coefficient is number $1$, which is a positive number. This means that this parabola will be “looking up”.

The $x$ coordinate of its vertex is:

$\frac{1 + (-2)}{2} = -\frac{1}{2}$ and $y$ coordinate is:

$ (\frac{1}{2})^2 – \frac{1}{2} – 2 = – \frac{9}{4}$

Now that we know all the data we need we can draw graph of this function.

Now let’s think a little bit what our inequality means. We are looking for all numbers for which this function’s values are positive numbers. Observe the graph- everything you need you can read right of it. You need intervals in which this graph is above $x$-axis.

Since given inequality is only greater than zero, zeros are not included in this solution. Solution to this inequality is:

$ < – \infty, -2 > \cup < 1, + \infty>$

Example 2. What if the previous task was given a bit differently and the inequality sign opposite?

$ x^2 + x – 2 < 0$

Once again:

$ x^2 + x – 2 = (x + 2)\cdot (x – 1)$

$$x_{1,2} = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}$$

$ x_1 = – 2$ and $ x_2 = 1$

The procedure of solving is the same, only now we are looking for values of x for which function values are negative – the interval where the graph is below $x$ – axis.

The solution to this inequality is the interval: $x\in < – 2, 1 >$.

Example 3. Solve the following inequality.

$ – x^2 – 3x – 4 > 0$

Again, first find the zeros, the vertex and then draw the graph. You’ll find that discriminant is lesser than zero (complex numbers) which means that this graph does not intersect the $x$- axis. This graph is faced downwards and is whole under the $x$- axis. So we do not have an set for this inequalities.

Example 4. Solve the following inequality.

$ – x^2 – 3x – 4 < 0$

Since now we are looking for the interval in which our graph is under the $x$- axis and it is completely under it, this means that the solution of this inequality is the whole set of real numbers.

Example 5. Solve the following inequality.

$\frac{2x – 1}{x – 3} \geq 1$

This is where you have to be careful because you can’t simply multiply this inequality with $x – 3$ and easily get rid of the fraction because you have no idea what this $x$ is and how it can affect inequality sign.

First thing you do is try to set this form of an inequality to a standard form $ b > 0$. This means that you should transfer everything to the left side and finding the common denominator.

$\frac{2x – 1}{x – 3} – 1 \geq 0$

$\frac{2x – 1 – x + 3}{x – 3} \geq 0$

$\frac{x + 2}{x – 3} \geq 0$

Now, you have to set your conditions. We only have one fraction with denominator is $x – 3$ which means that $ x \not= 3$.

This inequality is equivalent to $ (x + 2)(x – 3) \geq 0$ because quotient of two numbers is positive if and only if their product is also a positive number.

Zeros of this functions are $-2$ and $3$, and leading coefficient is a positive number which means that the solution to this inequality is: $ < – \infty, -2 > \cup < 3, + \infty>$.

Example 6. Solve the following inequality.

$\frac{(x – 2)(x + 2)}{(x – 1)(x + 3)} \le 0$

Now we got a negative fraction. This means that we can have two cases: first in which the numerator is negative and denominator is negative and second in which the numerator is positive and denominator negative. Be careful, the denominator can never be zero.

This inequality then is equivalent to two systems of quadratic inequalities.

a) $ (x – 2)(x + 2) \leq 0 , (x – 1)(x + 3) > 0$

b) $ (x – 2)(x + 2) \geq 0 , (x – 1)(x + 3) < 0$

The set of solutions is the union of solutions of these two systems. The easiest way to solve these inequalities is to draw the graphs and read the solutions of them.