# Quadratic Discriminant

The **discriminant** of a quadratic polynomial, denoted $\Delta,$ is a function of the coefficients of the polynomial, which provides information about the properties of the roots of the polynomial. By computing the discriminant, it is possible to distinguish whether the quadratic polynomial has two distinct real roots, one repeated real root, or non-real complex roots only.

Given a quadratic polynomial $ax^2+bx+c=0$ with real coefficients $a,b,$ and $c$ and $a \neq 0$, the discriminant of the polynomial is

$\Delta = b^2-4ac. \ _\square$

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## Explanation

From the quadratic formula, the roots of the quadratic polynomial $ax^2 + bx + c$ are given by

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.$

Now, observe that the discriminant is equal to the expression within the square root of the quadratic formula.

Since the quadratic formula gives all roots of the quadratic polynomial, we have the following cases:

- If $\Delta > 0 ,$ then the polynomial has two distinct real roots;
- If $\Delta = 0,$ then the polynomial has exactly one real root, which is a
**repeated root**; - If $\Delta < 0 ,$ then the expression inside the square root is negative and the roots are both non-real complex roots.

Also note that for a quadratic polynomial $ax^2 + bx + c ,$ if $b$ can be written in the form of $2b_0$ (where $b_0$ is also an integer), the quadratic formula is simplified as follows:

$x=\frac {-b_0 \pm \sqrt{{b_0}^2 -ac}}{a}.$

We often denote $\sqrt{{b_0}^2 -ac}$ in the above expression as $\Delta '$.

## Finding the Discriminant of a Quadratic

To compute the discriminant of a quadratic, we simply evaluate the expression $b^2 - 4ac$.

## Find the discriminant of the quadratic $x^2+3x+2$.

Since the discriminant of a quadratic polynomial $ax^2+bx+c=0$, where $a,b$ and $c$ are real numbers and $a \neq 0$, is $b^2-4ac$, the discriminant of the quadratic $x^2+3x+2$ is $3^2-4(1)(2)=1$. $_\square$

Note:In this case, there are two real roots of the quadratic polynomial.

## Find the discriminant of the quadratic polynomial $9x^2 + 6x + 1 =0.$

Again we can find the discriminant $\Delta$ as follows:

$\Delta = b^2 - 4\times a \times c = 6^2 - 4\times9\times 1 = 36 -36 = 0 . \ _\square$

Note:In this case, the discriminant is equal to zero, implying the quadratic polynomial has a repeated root.

## Repeated Roots

We have repeated roots when the discriminant is $0.$

## How many real roots are there to the quadratic polynomial $4x^2 + 4x + 1$?

Since the discriminant of a quadratic $ax^2+bx+c=0$, where $a,b$ and $c$ are real numbers and $a \neq 0$, is $b^2-4ac$, the discriminant of the quadratic $4x^2 + 4x + 1$ is $4^2 - 4\times4 = 0$. This indicates that the quadratic has a repeated root. Thus, there is $1$ real root. $_\square$

## Given that $2x^2 +8x + k$ has a repeated root, what is the value of $k$?

Since $\Delta = 0$ if there is a repeated root, we know that $8^2 - 4(2)(k) = 0.$ Simplifying, we get

$64 = 8k \Rightarrow k = 8. \ _\square$

## Range of Solutions

## For what values of $k$ does $-x^2 -kx - 5 =0$ have real solutions?

We know that there are only real solutions if the quadratic on the LHS has real roots, i.e. if $\Delta \geq 0$. Since $\Delta = k^2 - 4(-1)(-5) \geq 0$, we have

$k^2 \geq 20 \Rightarrow k \leq -\sqrt{20} \text{ or } k \geq \sqrt{20}.\ _\square$

## If the quadratic polynomial $x^2 + bx + c$ and the horizontal line $y = d$ intersect at exactly one point, what is the value of $d$?

To find the value of $x$ for which the polynomial and the line intersect, we equate the equations to obtain

$\begin{aligned} x^2 + bx + c &= d\\ x^2 + bx + (c-d) &= 0. \end{aligned}$

If there is exactly one intersection point, then this quadratic equation has exactly one solution. This occurs when the discriminant is zero, i.e.

$\Delta = b^2 - 4\cdot 1 \cdot (c-d) = b^2 - 4c + 4d = 0.$

Finally, we solve for $d$ to obtain $d = c - \frac{b^2}{4}.\ _\square$

Note:This agrees with the formula for computing the vertex of a parabola.

## Problem Solving

Be extremely careful when working on the following problem:

## Given a positive integer $m$, suppose the quadratic equation $x^2+mx+5m=0$ has two distinct real roots. What is the minimum possible value of $m?$

Since the quadratic equation has two distinct roots, the discriminant of the quadratic must be positive. Then

$\begin{aligned} \Delta=m^2-4\cdot 5m &>0\\ m^2-20m &>0 \\ m(m-20) &>0 \\ m&<0 \ \mbox{ or }\ m>20. \end{aligned}$

Since $m$ is positive, it must be the case that $m>20,$ and since $m$ is an integer, the minimum value of $m$ is $21.\ _\square$

## Using the discriminant, characterize the number of roots of the quadratic polynomial $x^2 + x + c$ as $c$ varies.

Computing the discriminant of the quadratic polynomial, we obtain

$\Delta = 1^2 - 4c = 1-4c.$

From the above characterization of roots using the discriminant, we have the following:

- If $1-4c > 0 ,$ then the polynomial has two distinct real roots. This occurs for $c< \frac{1}{4}$.
- If $1-4c = 0,$ then the polynomial has a repeated root. This occurs for $c=\frac{1}{4}$.
- If $1 -4c < 0 ,$ then the roots are both non-real complex roots. This occurs for $c> \frac{1}{4}$. $_\square$

## See Also

**Cite as:**Quadratic Discriminant.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/quadratic-discriminant/