# Quadratic inequalities

A quadratic inequality is a type of inequality in which a quadratic function appears. Quadratic inequalities have the form:

\( ax^2 + bx + c,,,,op,,,,0 \), where \(a\), \(b\) and \(c\) are real numbers and \(op\) is one of the inequality symbols \( < \), \( \le \), \(> \) or \( \ge \).

The solution to a quadratic inequality is the set of all values of \(x\) that satisfy the inequality. To find the solution set, we can use the following steps:

- Solve the corresponding quadratic equation \( ax^2 + bx + c = 0 \) to find the roots of the quadratic function.
- Plot the graph of the quadratic function on a coordinate plane.
- Determine the region(s) of the graph that are above or below the \(x\)-axis, depending on the sign of \(a\).
- Shade the appropriate region(s) of the graph, depending on the inequality symbol.
- Write the solution set using interval notation or set notation.

Let's look at some examples:

**Example 1:** Solve the inequality \(x^2 - 4x + 3 \ge 0 \).

Solution: First, we find the roots of the corresponding quadratic equation \(x^2 - 4x + 3 = 0 \) by factoring or using the quadratic formula:

\( x^2 - 4x + 3 = (x-1)(x-3)=0 \).

Thus, the roots are \( x=1 \) and \(x=3 \).

Next, we plot the graph of the quadratic function \( f(x) = x^2 - 4x + 3 \).

Graph

We can see that the graph of the function is above the \(x\)-axis between the roots \(x=1\) and \(x=3\).

Therefore, the solution to the inequality \( x^2 - 4x + 3 \ge 0 \) is: \( x \in (-\infty, 1] \cup [3, \infty) \).

**Example 2: An example of solving a quadratic inequality by the interval method.**

Let's consider the inequality: \( 2x^2 - 5x - 3 < 0 \).

To solve this inequality, we first find the roots of the quadratic equation \( 2x^2–5x–3=0 \)

\( x=- \frac{1}{2} \) and \( x = 3 \). We can find these roots by factoring the quadratic or by using the quadratic formula.

Next, we use the roots to divide the number line into three intervals:

\( (-\infty ; -\frac{1}{2} ) \), \( (-\frac{1}{2}; 3) \) and \( (3; \infty)\).

We then test a value from each interval in the inequality to determine whether it is true or false in that interval.

For the interval \( (-\infty ; -\frac{1}{2}) \) we can choose \( x= -1 \) as a test value.

Substituting \(x=-1\) into the inequality, we get: \( 2(-1)^2 - 5(-1) - 3 < 0 \), which simplifies to: \( 4 < 0 \).

This is false, so the inequality does not hold in the interval \( (-\infty ; -\frac{1}{2}) \) .

For the interval \( (-\frac{1}{2} ; 3) \) we can choose \(x=0\) as a test value.

Substituting \(x=0\) into the inequality, we get: \( 2(0)^2 - 5(0) - 3 < 0 \), which simplifies to: \( -3 < 0 \).

This is true, so the inequality holds in the interval \( (-\frac{1}{2} ; 3) \).

For the interval \( (3; \infty) \) we can choose \(x=4\) as a test value.

Substituting \(x=4\) into the inequality, we get: \(2(4)^2 - 5(4) - 3 < 0 \), which simplifies to: \( 5 < 0 \) .

This is false, so the inequality does not hold in the interval \( (3;\infty) \).

Therefore, the solution to the inequality \( 2x^2–5x–3 < 0 \) is: \( x\ \in (- \frac{1}{2} ; 3) \)

This means that any value of \(x\) that is greater than \(- \frac{1}{2} \) and less than \(3\) makes the inequality true.