# What is the discriminant of a quadratic function?

Jun 22, 2018

Below

#### Explanation:

The discriminant of a quadratic function is given by:

$\Delta = {b}^{2} - 4 a c$

What is the purpose of the discriminant?
Well, it is used to determine how many REAL solutions your quadratic function has

If $\Delta > 0$, then the function has 2 solutions
If $\Delta = 0$, then the function has only 1 solution and that solution is considered a double root
If $\Delta < 0$, then the function has no solution (you can't square root a negative number unless it's complex roots)

Aug 8, 2018

Given by the formula $\Delta = {b}^{2} - 4 a c$, this is a value computed from the coefficients of the quadratic that allows us to determine some things about the nature of its zeros...

#### Explanation:

Given a quadratic function in normal form:

$f \left(x\right) = a {x}^{2} + b x + c$

where $a , b , c$ are real numbers (typically integers or rational numbers) and $a \ne 0$, then the discriminant $\Delta$ of $f \left(x\right)$ is given by the formula:

$\Delta = {b}^{2} - 4 a c$

Assuming rational coefficients, the discriminant tells us several things about the zeros of $f \left(x\right) = a {x}^{2} + b x + c$:

• If $\Delta > 0$ is a perfect square then $f \left(x\right)$ has two distinct rational real zeros.

• If $\Delta > 0$ is not a perfect square then $f \left(x\right)$ has two distinct irrational real zeros.

• If $\Delta = 0$ then $f \left(x\right)$ has a repeated rational real zero (of multiplicity $2$).

• If $\Delta < 0$ then $f \left(x\right)$ has no real zeros. It has a complex conjugate pair of non-real zeros.

If the coefficients are real but not rational, the rationality of the zeros cannot be determined from the discriminant, but we still have:

• If $\Delta > 0$ then $f \left(x\right)$ has two distinct real zeros.

• If $\Delta = 0$ then $f \left(x\right)$ has a repeated real zero (of multiplicity $2$).

The discriminant occurs in the quadratic formula for the zeros of $a {x}^{2} + b x + c$, namely:

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a} = \frac{- b \pm \sqrt{\Delta}}{2 a}$

from which you can understand why the zeros have the nature they do for different values of $\Delta$.

Polynomials of higher degree also have discriminants, which when zero imply the existence of repeated zeros. The sign of the discriminant is less useful, except in the case of cubic polynomials, where it allows us to identify cases quite well...

Given:

$f \left(x\right) = a {x}^{3} + b {x}^{2} + c x + d$

with $a , b , c , d$ being real and $a \ne 0$.

The discriminant $\Delta$ of $f \left(x\right)$ is given by the formula:

$\Delta = {b}^{2} {c}^{2} - 4 a {c}^{3} - 4 {b}^{3} d - 27 {a}^{2} {d}^{2} + 18 a b c d$

• If $\Delta > 0$ then $f \left(x\right)$ has three distinct real zeros.

• If $\Delta = 0$ then $f \left(x\right)$ has either one real zero of multiplicity $3$ or two distinct real zeros, with one being of multiplicity $2$ and the other being of multiplicity $1$.

• If $\Delta < 0$ then $f \left(x\right)$ has one real zero and a complex conjugate pair of non-real zeros.