Discriminant of Quadratic Equation

Theorem

Let $\QQ$ be the quadratic equation:

$a x^2 + b x + c$

The discriminant of $\QQ$ is given by:

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

Proof

Let $\alpha_1$ and $\alpha_2$ be the roots of $\QQ$.

We have:

\(\ds \map \Delta \QQ\)

\(=\)

\(\ds \paren {\alpha_1 - \alpha_2}^2\)

Discriminant Example: Quadratic

\(\ds \)

\(=\)

\(\ds {\alpha_1}^2 - 2 \alpha_1 \alpha_2 + {\alpha_2}^2\)

Square of Difference

\(\ds \)

\(=\)

\(\ds {\alpha_1}^2 + 2 \alpha_1 \alpha_2 + {\alpha_2}^2 - 4 \alpha_1 \alpha_2\)

adding and subtracting $2 \alpha_1 \alpha_2$

\(\ds \)

\(=\)

\(\ds \paren {\alpha_1 + \alpha_2}^2 - 4 \alpha_1 \alpha_2\)

Square of Sum

\(\ds \)

\(=\)

\(\ds \paren {-\dfrac b a}^2 - 4 \dfrac c a\)

Sum of Roots of Quadratic Equation, Product of Roots of Quadratic Equation

\(\ds \)

\(=\)

\(\ds \dfrac {b^2 - 4 a c} {a^2}\)

simplification and common denominator

$\blacksquare$

Also presented as

While the discriminant $\Delta$ of a quadratic equation $\QQ: a x^2 + b x + c = 0$ is properly given as:

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

it is usual and practically universal that $\map \Delta \QQ$ is given as:

$\map \Delta \QQ = b^2 - 4 a c$

This is because we are usually interested only in the sign of $\map \Delta \QQ$ and not its actual value.

Indeed, we note that as $a^2 > 0$, multiplying or dividing by $a^2$ has no effect on that sign.

Sources

• 1998: David Nelson: The Penguin Dictionary of Mathematics (2nd ed.) ... (previous) ... (next): discriminant (of a polynomial equation)
• 2008: David Nelson: The Penguin Dictionary of Mathematics (4th ed.) ... (previous) ... (next): discriminant (of a polynomial equation)