# DETERMINE THE NATURE OF THE SOLUTIONS OF THE EQUATION

Use The Quadratic Formula Calculator to watch Quadratic Formula and discriminant in Action! This calculator will certainly settle any kind of quadratic equation you form in (even if options are imaginary).

To understand what the discriminant does, it"s necessary that you have actually a good knowledge of:

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

A parabola.

The solution can be thought of in two different ways.

Algebraically, the solution occurs when y = 0. So the solution is wright here \$\$y = ed ax^2 + lue bx + colorgreen c \$\$ becomes \$\$0 = ed ax^2 + lue bx + colorgreen c \$\$. Graphically, since y = 0 is the x-axis, the solution is wbelow the parabola intercepts the x-axis. (This only functions for real solutions).

In the photo listed below, the left parabola has 2 actual remedies (red dots), the middle parabola has 1 actual solution (red dot) and also the best the majority of parabola has actually no actual solutions (yes, it does have actually imaginary ones).

What does the discriminant look like?

It looks favor .. a number.

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

What is the discriminant anyway?

The discriminant is a number that deserve to be calculated from any kind of quadratic equation.

A quadratic equation is an equation that deserve to be composed as \$\$ ax^2 + bx + c \$\$ (where \$\$a e 0 \$\$).

What is the formula for the Discriminant?

The discriminant for any type of quadratic equation of the create \$\$ y = ed a x^2 + lue bx + shade green c \$\$ is found by the following formula and also it provides critical information regarding the nature of the roots/services of any kind of quadratic equation.

\$oxedFormula\ extDiscriminant = lue b^2 -4 ed a colorgreen c\$

\$oxedExample\ extEquation : y = ed 3 x^2 + lue 9x + color green 5\ extDiscriminant = lue 9^2 -4 cdot ed 3 cdot colorgreen 5 \ extDiscriminant =oxed 6\$

What does this formula tell us?

The discriminant tells us the adhering to information around a quadratic equation:

If the solution is one unique number or 2 various numbers.

### Nature of the Solutions

Value of the discriminant Type and variety of Solutions Example of graph

\$ b^2 - 4ac > 0 \$

\$ extExample :\y = ed 3x^2 lue-6x + colorgreen 2 \ extDiscriminant\lue-6^2 - 4 cdot ed 3 cdot colorgreen 2\= oxed12\\$

If the discriminant is a positive number, then tbelow are 2 actual options. This implies that the graph of the parabola interepts the x-axis at 2 unique points .   \$ extExample :\y = ed 3x^2 + lue 4 x colorgreen -4 \ extDiscriminant\lue 4^2 - 4 cdot ed 3 cdot colorgreen -4\= oxed64\\$

If the discriminant is positive and also a perfect square choose 64, then tright here are 2 genuine rational solutions.

\$ extExample :\y = ed 3x^2 lue -6 x + colorgreen 2 \ extDiscriminant\lue -6^2 - 4 cdot ed 3 cdot colorgreen 2\= oxed12 \\$

If the discriminant is positive and also not a perfect square choose 12, then tbelow are 2 real irrational remedies.

 \$ b^2 - 4ac = 0 \$\$ extExample :\y = ed 4 x^2 lue-28x + colorgreen 49 \ extDiscriminant\lue-28^2 - 4 cdot ed 4 cdot colorgreen 49\= oxed0\\$ \$ b^2 - 4ac Tright here are just imaginary Solutions. This means that the graph of the quadratic never before intersects the axes.

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

\$a = ed 1\b = lue 2\a = colorgreen 1\$

The discriminant for this equation is:

\$ extDiscriminant = lue b^2 -4 ed a colorgreen c\ extDiscriminant = lue 2^2 -4 cdot ed 1 cdot colorgreen cdot 1\ extDiscriminant = 4 -4\ extDiscriminant = oxed0\ \$

Due to the fact that the discriminant is zero, tright here need to be 1 genuine solution to this equation. Below is a photo representing the graph and the one solution of \$\$ y = x^2 + 2x + 1\$\$.

In this quadratic equation,\$\$ y = ed 1 x^2 + lue -2 + color green 1 \$\$

\$ extEquation : y = ed 1 x^2 + lue -2x + shade green 1\a = ed 1\b = lue-2\c = colorgreen 1\$

Using our basic formula:

\$\$ extDiscriminant \eginaligned&= lue b^2 -4 cdot ed a cdot colorgreen c \&= lue -2^2 -4 cdot ed 1 cdot colorgreen 1 \&= oxed0endaligned \$\$

Due to the fact that the discriminant is zero, we have to suppose 1 real solution which you deserve to view pictured in the graph below.

In this quadratic equation,\$\$ y = ed 1 x^2 + lue -1x + shade green 1 \$\$

\$ extEquation : y = ed 1 x^2 + lue -1 + color green 1\a = ed 1\b = lue-1\c = colorgreen 2\$

Using our basic formula:

\$\$ extDiscriminant \eginaligned&= lue b^2 -4 cdot ed a cdot colorgreen c \&= lue -1^2 -4 cdot ed 1 cdot colorgreen -2 \&= 1 - -8 \ &= 1 + 8 = oxed 9endaligned \$\$

Because the discriminant is positive and also rational, tright here have to be 2 actual rational solutions to this equation. As you can see listed below, if you usage the quadratic formula to discover the actual remedies, you do indeed get 2 genuine rational remedies.