# Which Function Has a Negative Discriminant Value

Which Function Has a Negative Discriminant Value.

Utilize The Quadratic Formula Calculator to see Quadratic Formula and discriminant in Action! This calculator will solve any quadratic equation you type in
(even if solutions are imaginary).

To understand what the discriminant does, it’southward important that y’all have a good understanding of:

#### Pre Req two : What is the solution of a quadratic equation:

The solution can be idea of in two dissimilar means.

1. Algebraically, the solution occurs when y = 0. Then the solution is where
$$y =\cherry-red ax^two + \blue bx + \color{greenish} c$$
becomes
$$0 =\red ax^2 + \bluish bx + \color{green} c$$.
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 existent solution (red dot) and the right nigh parabola has no real solutions (yes, information technology does have imaginary ones).

#### What does the discriminant look like?

It looks like .. a number.

5, 2, 0, -one
– each of these numbers is the discriminant for iv different quadratic 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 equally $$ax^ii + bx + c$$
(where $$a \ne 0$$).

#### What is the formula for the Discriminant?

The discriminant for any quadratic equation of the form
$$y =\red a 10^ii + \blue bx + \color {green} c$$

is found past the following formula and it provides critical information regarding the nature of the roots/solutions of whatever quadratic equation.

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

$\boxed{Example} \\ \text{Equation :} y =\red 3 10^2 + \blueish 9x + \colour {light-green} 5 \\ \text{Discriminant } = \blue ix^2 -4 \cdot \red 3 \cdot \color{green} 5 \\ \text{Discriminant } = \boxed{ six}$

#### What does this formula tell usa?

Respond

The discriminant tells us the following information nigh 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 2 different numbers.

### Nature of the Solutions

Value of the discriminant Blazon and number of Solutions Example of graph

$b^2 – 4ac > 0$

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

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 singled-out points .

$\text{Example :} \\ y = \carmine 3x^2 + \bluish 4 x \color{green} {-4} \\ \text{Discriminant} \\ \blue four^2 – 4 \cdot \crimson 3 \cdot \colour{green} {-4} \\ = \boxed{64} \\$

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

$\text{Example :} \\ y = \ruby-red 3x^2 \blue {-6} x + \colour{green} 2 \\ \text{Discriminant} \\ \blue {-6}^2 – iv \cdot \ruby iii \cdot \color{green} ii \\ = \boxed{12 } \\$

If the discriminant is positive and
not
a perfect foursquare similar 12, and then there are 2 real

irrational

solutions.

 $b^2 – 4ac = 0$ $\text{Example :} \\ y = \red 4 x^2 \blueish{-28}x + \colour{light-green} {49} \\ \text{Discriminant} \\ \blue{-28}^ii – iv \cdot \red 4 \cdot \colour{green} 49 \\ = \boxed{0} \\$ $b^2 – 4ac < 0$ $\text{Example :} \\ y = \red 10^2 \blue{-three}x + \color{light-green} 4 \\ \text{Discriminant} \\ \blue{-iii}^2 – iv \cdot \red i \cdot \color{green} iv \\ = \boxed{-7} \\$ There are only imaginary Solutions. This means that the graph of the quadratic never intersects the axes.

### Case

$$y = x^2 + 2x + one$$

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

The discriminant for this equation is:

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

Since the discriminant is zero, at that place should be 1 real solution to this equation.

Below is a picture representing the graph and the one solution of

$$y = x^2 + 2x + 1$$
.

### Practice Problems

##### Practice i

$$y =\blood-red 1 x^ii + \blue {-2} + \colour {green} 1$$

$\text{Equation : } y =\cerise i x^2 + \blueish {-2}10 + \colour {green} 1 \\ a = \scarlet 1 \\ b = \blueish{-ii} \\ c = \color{green} 1$

Using our general formula:

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

Since the discriminant is aught, we should look 1 existent solution which you can see pictured in the graph beneath.

##### Practice two

$$y =\red 1 x^2 + \blue {-1}10 + \color {green} 1$$

$\text{Equation : } y =\red 1 ten^2 + \blueish {-one} + \colour {green} 1 \\ a = \ruddy ane \\ b = \blue{-1} \\ c = \color{greenish} {2}$

Using our general formula:

\text{Discriminant } \\ \brainstorm{aligned} &= \blue b^2 -four \cdot \red a \cdot \color{green} c \\ &= \blue {-i}^2 -4 \cdot \red 1 \cdot \color{dark-green} {-two} \\ &= i – -8 \\ &= 1 + 8 = \boxed 9 \end{aligned}

Since the discriminant is positive and rational, in that location should be 2 real rational solutions to this equation. Every bit you lot can see beneath, if you apply the quadratic formula to notice the actual solutions, you do indeed get two real rational solutions.

##### Do 3

y =
one10²
− 1
.

• a = i
• b = 0
• c = − 1

$$\colour{Crimson}{b^two} – 4\colour{Magenta}{a}\color{Blue}{c} \\ \color{Ruby-red}{(0)^two} – four\color{Magenta}{(1)}\colour{Blue}{(-1)} = iv$$

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

Again if yous’d like to come across the bodily solutions and the graph, merely wait below:

##### Practise four

y = x² + 4x − 5.

• a = 1
• b = 4
• c = − v

$$\color{Red}{b^2} – 4\colour{Magenta}{a}\color{Blue}{c} \\ \color{Red}{(4)^2} – 4\colour{Magenta}{(one)}\color{Blue}{(-5)} \\ xvi – iv(-5) = 16 +twenty \\ = 36$$

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

##### Practise 5

y = ten² – 4x + 5.

• a = one
• b = -4
• c = v

$$\colour{Red}{b^2} – 4\color{Magenta}{a}\colour{Bluish}{c} \\ \color{Ruddy}{(-4)^2} – 4\color{Magenta}{(1)}\color{Blueish}{(5)} \\ = 16 – 20 = -4$$

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

Below is a picture of this quadratic’s graph.

##### Practice half dozen

y = x² + 4

• a = one
• b = 0
• c = 4

$$\color{Red}{b^2} – 4\color{Magenta}{a}\color{Blue}{c} \\ \color{Scarlet}{(0)^two} – 4\colour{Magenta}{(i)}\color{Blue}{(four)} = -sixteen$$

Since the discriminant is negative, in that location are two imaginary solutions to this quadratic equation.

The solutions are
2i
and
-2i.

Below is a picture of this equations graph.

##### Practice 7

y = x² + 25

• a = ane
• b = 0
• c = 25

$$\color{Red}{b^2} – 4\colour{Magenta}{a}\color{Bluish}{c} \\ \color{Crimson}{(0)^two} – iv\color{Magenta}{(1)}\color{Blue}{(25)} = -100$$

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

The solutions are
5i
and
-5i.