# How To Solve Quadratic Inequalities?

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It’s quite easy to solve linear inequalities of the type $5x – 3 > 0$. But do you know how to solve quadratic inequalities of the type $5x^{2} – 3x + 8 \gt 0$?

## Difference Between Equation and Inequation (Inequality)

An equation is a statement that maintains the equal value of two mathematical expressions. An inequality, on the other hand, is a statement that uses the symbols $\gt$ for greater than or $\lt$ for lesser than to denote that one quantity is larger or smaller in value than another.

For example, $5x – 7 = 0$ is an equation, whereas $5x – 7 \lt 0$ is an inequality.

Following are the main differences between an equation and an inequality.

• An equation uses a symbol $=$ while an inequality uses symbols such as $\lt$ and $\gt$.
• An equation is a mathematical statement that shows the equal value of two expressions while an inequality is a mathematical statement that shows that an expression is lesser than or more than the other.
• An equation shows the equality of two variables while an inequality shows the inequality of two variables.
• Although both can have several different solutions, an equation only has one answer while an inequality has several answers.

## What is a Quadratic Inequality?

A quadratic equation is an algebraic equation of the second degree in $x$. The quadratic equation in its standard form is $ax^{2} + bx + c = 0$, where $a$ and $b$ are the coefficients, $x$ is the variable, and $c$ is the constant term. The condition for an equation to be a quadratic equation is that the coefficient of $x^{2}$ is a non-zero term$\left(a \ne 0 \right)$.

If in a quadratic equation, equality sign ‘$=$’ is replaced by inequality sign ‘$\lt$’, ‘$\gt$’, ‘$\le$’ or ‘$\ge$’, then it becomes a quadratic inequality.

The general form of a quadratic inequality is either of the following:

• $ax^{2} + bx + c \lt 0$
• $ax^{2} + bx + c \le 0$
• $ax^{2} + bx + c \gt 0$
• $ax^{2} + bx + c \ge 0$

## How to Solve Quadratic Inequalities?

To understand how to solve quadratic inequalities, let’s consider the following example.

$x^{2} – 5x + 6 \lt 0$

The first step in solving a quadratic inequality is to convert it to its corresponding equation.

Corresponding equation of $x^{2} – 5x + 6 \lt 0$ is $x^{2} – 5x + 6 = 0$.

Now solve this quadratic equation. You can use any of the methods to solve it – factorization (splitting the middle term), completing the square or using quadratic formula.

Let’s use splitting the middle term method.

$x^{2} – 5x + 6 = 0 => x^{2} – 2x – 3x + 6 = 0 => x \left( x – 2 \right) – 3 \left( x – 2 \right) = 0 => \left( x – 2 \right) \left( x – 3 \right) = 0$

$=> x – 2 = 0$ or $x – 3 = 0$ $=> x = 2$ or $x = 3$.

So, the two solutions of the quadratic equation $x^{2} – 5x + 6 = 0$ are $x = 2$ and $x = 3$.

Now draw a number line and plot these two points ($x = 2$ and $x = 3$).

You can see that the points $x = 2$ and $x = 3$ divides the number line in three regions:

• Region 1: $\left( – \infty, 2 \right)$
• Region 2: $\left(2, 3 \right)$
• Region 3: $\left( 3, \infty \right)$

Next consider a random point in each of these regions. (You can choose any points, but make sure they do not lead to complex or lengthy calculations).

Let’s choose the following points:

• Region 1: $x = 0$ for $\left( – \infty, 2 \right)$
• Region 2: $x = 2.5$ for $\left(2, 3 \right)$
• Region 3: $x = 4$ for $\left( 3, \infty \right)$ Regions on a number line created by points $x = 0$, $x = 2.5$ and $x = 3$

Now substitute each of these points in the given quadratic inequality one by one.

Substituting $x = 0$ in $x^{2} – 5x + 6 \lt 0 => 0^{2} – 5 \times 0 + 6 \lt 0 => 6 \lt 0$ (False). Hence, $\left( – \infty, 2 \right)$ cannot be the solution.

Substituting $x = 2.5$ in $x^{2} – 5x + 6 \lt 0 => 2.5^{2} – 5 \times 2.5 + 6 \lt 0 => -0.25 \lt 0$ (True). Hence, $\left(2, 3 \right)$ is the solution.

Substituting $x = 4$ in $x^{2} – 5x + 6 \lt 0 => 4^{2} – 5 \times 4 + 6 \lt 0 => 2 \lt 0$ (False). Hence, $\left( 3, \infty \right)$ cannot be the solution. Graphical solution of $x^{2} – 5x + 6 \lt 0$

Therefore, the solution of $x^{2} – 5x + 6 \lt 0$ is $\left(2, 3 \right)$.

## Let’s Code With Python

# Accept Data
print('For Quadratic Equation of the form ax^2 + bx + c = 0, enter for a, b & c')
a = float(input('Enter value for a '))
b = float(input('Enter value for b '))
c = float(input('Enter value for c '))
print('Quadratic equation is ', a, 'x^2 +', b, 'x +', c)

# Import Libraries
import numpy as np
import matplotlib.pyplot as plt

#Define Data
x = np.arange(-10, 10, 0.2)
y = a * (x ** 2) + b * x + c
ymin = 0
ymax = 0

# Define x and y Range
for i in range(-10, 10, 1):
y1 = a * (i ** 2) + b * i + c
if y1 < ymin:
ymin = y1
if y1 > ymax:
ymax = y1

# Plot
plt.plot(x, y)

# Set axes limit
plt.xlim(-12, 12)
plt.ylim(ymin - 5,ymax + 5)

#plt.title('Quadratic equation', a, 'x^2 +', b, 'x +', c)

plt.xlabel("X-axis")
plt.ylabel("Y-axis")

# Display
plt.show()

## Conclusion

A quadratic inequality is not solved by using the normal method as used with linear inequalities. In order to solve a quadratic inequality, the first step is to find the solutions of corresponding quadratic equation and then these solutions are used to solve the quadratic inequality.

## Practice Problems

• $x^{2} – 8x + 16 \lt 0$
• $x^{2} + 4x + 4 \gt 0$
• $x^{2}+4x – 5 \le 0$
• $2x^{2} – 5x + 3 \ge = 0$
• $x^{2} + 4x + 5 \ge = 0$
• $2x^{2} + x – 300 \le = 0$
• $3x^{2} -2x + \frac{1}{3} \lt 0$
• $2x^{2} + x – 528 \gt 0$