﻿ The Discriminant in Quadratic Equations--visual tutorial with examples, practice problems and free printable pdf

Nature of Quadratic Equation's roots, solutions

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What does the discriminant look like?

It looks like .. a number

5,2, 0 -1 - this is the discriminant for 4 different equations

What is the discriminant anyway?

The discriminant is a number that can be calculated from any quadratic equation

A quadratic equation is an equation that can be written as $$ax^2 +bx + c$$ ( where $$a \ne 0$$)

What is the formula for the Discriminant?

The discriminant in a quadratic equation is found by the following formula and the discriminant provides critical information regarding the nature of the roots/solutions of any quadratic equation.

discriminant= b² − 4ac

Example of the discriminant
• Quadratic equation = y = 3x² + 9x + 5
• The discriminant = 9 ² − 4 • 3 •5

Important pre requisites

To understand what the discriminant does, it's important that you have a good understanding of

What is the solution of a quadratic equation:

The solution can be thought of in two different ways.

1. Algebraically, the solution occurs when y = 0. So the solution is where $$y =\color{Magenta}ax^2 + \red bx + \color{Blue}{c}$$ becomes $$0 =\color{Magenta}ax^2 + \red bx + \color{Blue}{c}$$

2. Graphically, since y = 0 is the x-axis, the solution is where the parabola intercepts the x-axis. (This only works for real solutions).

In the picture below, the left parabola has 2 real solutions (red dots), the middle parabola has 1 real solution (red dot) and the right most parabola has no real solutions (yes, it does have imaginary ones)

What does this formula tell us?

• If the solution is a real number or an imaginary number.
• If the solution is rational or if it is irrational.
• If the solution is 1 unique number or two different numbers

Nature of the Solutions

 Value of the discriminant Type and number of Solutions Example of graph Positive Discriminant b² − 4ac > 0 Two Real SolutionsIf the discriminant is a perfect square the roots are rational. Otherwise, they are irrational. Positive and a Perfect Square b² − 4ac = perfect square Two Real Rational Solutions. Positive and a not a perfect square b² − 4ac = not a perfect square Two Real Irrational Solutions. Discriminant is Zero b² − 4ac = 0 One Real Solution Negative Discriminant b² − 4ac < 0 No Real Solutions Two Imaginary Solutions

Example

Quadratic Equation: y = x² + 2x + 1

• a = 1
• b = 2
• c = 1

The discriminant for this equation is

$$\color{Red}{b^2} - 4\color{Magenta}{a}\color{Blue}{c} \\ \color{Red}{2^2} - 4\color{Magenta}{(1)}\color{Blue}{(1)} = 0$$

Since the discriminant is zero, there should be 1 real solution to this equation.
Below is a picture representing the graph and one solution of this quadratic equation Graph of y = x² + 2x + 1

Practice Problems

In this quadratic equation,y = 1 − 2x + 1

• a = 1
• b = − 2
• c = 1

Using our general formula, the discriminant is

$$\color{Red}{b^2} - 4\color{Magenta}{a}\color{Blue}{c} \\ \color{Red}{(-2)^2} - 4\color{Magenta}{(1)}\color{Blue}{(1)} = 4$$

Since the discriminant is zero, we should expect 1 real solution which you can see pictured in the graph below.

In this quadratic equation, y = x² − x − 2 and its solution

• a = 1
• b = − 1
• c = − 2

Discriminant

$$\color{Red}{b^2} - 4\color{Magenta}{a}\color{Blue}{c} \\ \color{Red}{(-1)^2} - 4\color{Magenta}{(1)}\color{Blue}{(-2)} \\ 1 - -8 = 1 + 8 = 9$$

Since the discriminant is positive and rational , there should be 2 real rational solutions to this equation. As you can see below, if you use the quadratic formula to find the actual solutions, you do indeed get 2 real rational solutions.

In this quadratic equation, y = 1− 1

• a = 1
• b = 0
• c = − 1

$$\color{Red}{b^2} - 4\color{Magenta}{a}\color{Blue}{c} \\ \color{Red}{(0)^2} - 4\color{Magenta}{(1)}\color{Blue}{(-1)} = 4$$

Since the discriminant is positive and a perfect square, we have two real solutions that are rational.

Again if you'd like to see the actual solutions and the graph, just look below:

In this quadratic equation, y = x² + 4x − 5

• a = 1
• b = 4
• c = − 5

$$\color{Red}{b^2} - 4\color{Magenta}{a}\color{Blue}{c} \\ \color{Red}{(4)^2} - 4\color{Magenta}{(1)}\color{Blue}{(-5)} \\ 16 - 4(-5) = 16 +20 \\ = 36$$

Since this quadratic equation's discriminant is positive and a perfect square, there are two real solutions that are rational.

In this quadratic equation, y = x² - 4x + 5

• a = 1
• b = -4
• c = 5

$$\color{Red}{b^2} - 4\color{Magenta}{a}\color{Blue}{c} \\ \color{Red}{(-4)^2} - 4\color{Magenta}{(1)}\color{Blue}{(5)} \\ = 16 - 20 = -4$$

Since the discriminant is negative , there are no real solutions to this quadratic equation. The only solutions are imaginary.

Below is a picture of this quadratic's graph

y = x² + 4

• a = 1
• b = 0
• c = 4

$$\color{Red}{b^2} - 4\color{Magenta}{a}\color{Blue}{c} \\ \color{Red}{(0)^2} - 4\color{Magenta}{(1)}\color{Blue}{(4)} = -16$$

Since the discriminant is negative , there are two imaginary solutions to this quadratic equation.

The solutions are 2i and -2i

Below is a a picture of this equations graph

y = x² + 25

• a = 1
• b = 0
• c = 25

$$\color{Red}{b^2} - 4\color{Magenta}{a}\color{Blue}{c} \\ \color{Red}{(0)^2} - 4\color{Magenta}{(1)}\color{Blue}{(25)} = -100$$

Since the discriminant is negative , there are two imaginary solutions to this quadratic equation.

The solutions are 5i and -5i

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