How To Solve Quadratic Inequalities?

This post is also available in: हिन्दी (Hindi)

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 $5x^{2} – 3x + 8 \gt 0$?

Solving quadratic inequalities involve two steps. In this article, you will learn how to solve quadratic inequalities.

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)$

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.

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

Let’s Code With Python

Plot a quadratic function.

# 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)

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

# Add labels
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

Solve the following quadratic inequalities:

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

Recommended Reading

• Which One Will You Choose: Two 5-Inch Pizzas Or One 9-Inch Pizza?
• Difference Between Common Logarithm and Natural Logarithm
• What Does e(Euler Constant) Mean in Math?

Image Credit: School stationery vector created by upklyak – www.freepik.com