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Exercise 1110Algebra › PolynomialsL1Added 3 Nov 2026

The discriminant of a quadratic

#discriminant#quadratic#roots

Statement

Express the discriminant of aX2+bX+caX^2+bX+c in terms of the roots, and connect it to the nature of the roots.

Hint

Δ=b2−4ac=a2(x1−x2)2\Delta=b^2-4ac=a^2(x_1-x_2)^2.

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Solution

Relation to roots. By Vieta, x1+x2=−b/ax_1+x_2=-b/a, x1x2=c/ax_1x_2=c/a. Then
(x1−x2)2=(x1+x2)2−4x1x2=b2a2−4ca=b2−4aca2=Δa2.(x_1-x_2)^2=(x_1+x_2)^2-4x_1x_2=\frac{b^2}{a^2}-\frac{4c}{a}=\frac{b^2-4ac}{a^2}=\frac{\Delta}{a^2}.
So Δ=a2(x1−x2)2\Delta=a^2(x_1-x_2)^2.

Sign of Δ\Delta.
- If Δ>0\Delta>0: two distinct real roots.
- If Δ=0\Delta=0: one double real root.
- If Δ<0\Delta<0: two complex conjugate roots.

Conclusion. The sign of Δ\Delta determines the nature of the roots, and ∣Δ∣1/2=∣a∣∣x1−x2∣|\Delta|^{1/2}=|a||x_1-x_2| is the distance between the roots (in R\mathbb{R}).