Find the Discriminant and Number of Real Solutions

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Pre Requisite Knowledge

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

Pre Req 1 : What does the graph of a quadratic equation look like:

Pre Req 2 : What is the solution of a quadratic equation:

Answer

The solution can be thought of in two different ways.

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

x intercept

What does the discriminant look like?

Answer

It looks like .. a number.

5, 2, 0, -1 - each of these numbers is the discriminant for 4 different quadratic equations.

What is the discriminant anyway?

Answer

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?

Answer

The discriminant for any quadratic equation of the form $$ y =\red a x^2 + \blue bx + \color {green} c $$ is found by the following formula and it provides critical information regarding the nature of the roots/solutions of any quadratic equation.

$ \boxed{Formula} \\ \text{Discriminant } = \blue b^2 -4 \red a \color{green} c $

$ \boxed{Example} \\ \text{Equation :} y =\red 3 x^2 + \blue 9x + \color {green} 5 \\ \text{Discriminant } = \blue 9^2 -4 \cdot \red 3 \cdot \color{green} 5 \\ \text{Discriminant } = \boxed{ 6} $

What does this formula tell us?

Answer

The discriminant tells us the following information about a quadratic equation:

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 one unique number or two different numbers.

Nature of the Solutions

Value of the discriminant	Type and number of Solutions	Example of graph
$ b^2 - 4ac > 0 $ $ \text{Example :} \\ y = \red 3x^2 \blue{-6}x + \color{green} 2 \\ \text{Discriminant} \\ \blue{-6}^2 - 4 \cdot \red 3 \cdot \color{green} 2 \\ = \boxed{12} \\ $	If the discriminant is a positive number, then there are 2 real solutions. This means that the graph of the parabola interepts the x-axis at 2 distinct points .
$ \text{Example :} \\ y = \red 3x^2 + \blue 4 x \color{green} {-4} \\ \text{Discriminant} \\ \blue 4^2 - 4 \cdot \red 3 \cdot \color{green} {-4} \\ = \boxed{64} \\ $	If the discriminant is positive and also a perfect square like 64, then there are 2 real rational solutions.
$ \text{Example :} \\ y = \red 3x^2 \blue {-6} x + \color{green} 2 \\ \text{Discriminant} \\ \blue {-6}^2 - 4 \cdot \red 3 \cdot \color{green} 2 \\ = \boxed{12 } \\ $	If the discriminant is positive and not a perfect square like 12, then there are 2 real irrational solutions.

Value of the discriminant

Type and number of Solutions

Example of graph

Positive Discriminant Title

$ b^2 - 4ac > 0 $

$ \text{Example :} \\ y = \red 3x^2 \blue{-6}x + \color{green} 2 \\ \text{Discriminant} \\ \blue{-6}^2 - 4 \cdot \red 3 \cdot \color{green} 2 \\ = \boxed{12} \\ $

If the discriminant is a positive number, then there are 2 real solutions. This means that the graph of the parabola interepts the x-axis at 2 distinct points .

Positive and perfect square Discriminant Title

$ \text{Example :} \\ y = \red 3x^2 + \blue 4 x \color{green} {-4} \\ \text{Discriminant} \\ \blue 4^2 - 4 \cdot \red 3 \cdot \color{green} {-4} \\ = \boxed{64} \\ $

If the discriminant is positive and also a perfect square like 64, then there are 2 real rational solutions.

picture of graph of parabols with a positive discriminant

Positive and perfect square Discriminant Title

$ \text{Example :} \\ y = \red 3x^2 \blue {-6} x + \color{green} 2 \\ \text{Discriminant} \\ \blue {-6}^2 - 4 \cdot \red 3 \cdot \color{green} 2 \\ = \boxed{12 } \\ $

If the discriminant is positive and not a perfect square like 12, then there are 2 real irrational solutions.

$ b^2 - 4ac = 0 $

$ \text{Example :} \\ y = \red 4 x^2 \blue{-28}x + \color{green} {49} \\ \text{Discriminant} \\ \blue{-28}^2 - 4 \cdot \red 4 \cdot \color{green} 49 \\ = \boxed{0} \\ $

$ b^2 - 4ac < 0 $

$ \text{Example :} \\ y = \red x^2 \blue{-3}x + \color{green} 4 \\ \text{Discriminant} \\ \blue{-3}^2 - 4 \cdot \red 1 \cdot \color{green} 4 \\ = \boxed{-7} \\ $

There are only imaginary Solutions. This means that the graph of the quadratic never intersects the axes.

Example

Quadratic Equation: $$ y = x^2 + 2x + 1$$

$ a = \red 1 \\ b = \blue 2 \\ a = \color{green} {1} $

The discriminant for this equation is:

$ \text{Discriminant } = \blue b^2 -4 \red a \color{green} c \\ \text{Discriminant } = \blue 2^2 -4 \cdot \red 1 \cdot \color{green} \cdot 1 \\ \text{Discriminant } = 4 -4 \\ \text{Discriminant } = \boxed{0} \\ $

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

Picture of graph of solved quadratic formula

Practice Problems

Practice 1

Calculate the discriminant to determine the number and nature of the solutions of the following quadratic equation: $$y = x² − 2x + 1 $$.

In this quadratic equation, $$ y =\red 1 x^2 + \blue {-2} + \color {green} 1 $$

$ \text{Equation : } y =\red 1 x^2 + \blue {-2}x + \color {green} 1 \\ a = \red 1 \\ b = \blue{-2} \\ c = \color{green} 1 $

Using our general formula:

$$ \text{Discriminant } \\ \begin{aligned} &= \blue b^2 -4 \cdot \red a \cdot \color{green} c \\ &= \blue {-2}^2 -4 \cdot \red 1 \cdot \color{green} 1 \\ &= \boxed{0} \end{aligned} $$

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

solution 1 by 0 graph

Practice 2

Use the discriminant to find out the nature and number of solutions: $$y = x² − x − 2 $$.

In this quadratic equation, $$ y =\red 1 x^2 + \blue {-1}x + \color {green} 1 $$

$ \text{Equation : } y =\red 1 x^2 + \blue {-1} + \color {green} 1 \\ a = \red 1 \\ b = \blue{-1} \\ c = \color{green} {2} $

Using our general formula:

$$ \text{Discriminant } \\ \begin{aligned} &= \blue b^2 -4 \cdot \red a \cdot \color{green} c \\ &= \blue {-1}^2 -4 \cdot \red 1 \cdot \color{green} {-2} \\ &= 1 - -8 \\ &= 1 + 8 = \boxed 9 \end{aligned} $$

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.

solution 2 and 1 solution 2 and 1 graph

Practice 3

Calculate the discriminant to determine the nature and number of solutions: y = x² − 1.

In this quadratic equation, y = 1x² − 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:

solution 1 and -1 solution 1 and -1 graph

Practice 4

Calculate the discriminant to determine the nature and number of solutions: y = x² + 4x − 5.

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.

solution 5 and -1

Practice 5

Calculate the discriminant to determine the nature and number of solutions: y = x² - 4x + 5.

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.

formula imaginary 1

Below is a picture of this quadratic's graph.

solutions imaginary 1

Practice 6

Find the discriminant to determine the nature and number of solutions: y = x² + 4 .

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 picture of this equations graph.

solutions imaginary 2

Practice 7

Find the discriminant to determine the nature and number of solutions: y = x² + 25.

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.

Find the Discriminant and Number of Real Solutions

Source: https://www.mathwarehouse.com/quadratic/discriminant-in-quadratic-equation.php

Prerauer Monsond

Find the Discriminant and Number of Real Solutions

Pre Requisite Knowledge

Pre Req 1 : What does the graph of a quadratic equation look like:

Pre Req 2 : What is the solution of a quadratic equation:

What does the discriminant look like?

What is the discriminant anyway?

What is the formula for the Discriminant?

What does this formula tell us?

Nature of the Solutions

Example

Practice Problems

Practice 1

Practice 2

Practice 3

Practice 4

Practice 5

Practice 6

Practice 7

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