If the quartic is monic (a = 1), the sum of squares α^2 + β^2 + γ^2 + δ^2 equals which of the following?

If the quartic is monic (a = 1), the sum of squares α^2 + β^2 + γ^2 + δ^2 equals which of the following?

• A. b^2 - 2c
• B. b^2 + 2c
• C. -b^2 + 2c
• D. -b^2 - 2c

Answer: (A)

The key idea is to relate the sum of the squares of the roots to the sums of the roots and their pairwise products, using Vieta’s formulas for a monic quartic. Let the roots be α, β, γ, δ. Define S1 = α+β+γ+δ and S2 = αβ + αγ + αδ + βγ + βδ + γδ. A monic quartic with these roots has the form x^4 − S1 x^3 + S2 x^2 − S3 x + S4, so the coefficients give S1 = −b and S2 = c. The sum of squares can be written as α^2+β^2+γ^2+δ^2 = (α+β+γ+δ)^2 − 2(αβ+αγ+αδ+βγ+βδ+γδ) = S1^2 − 2S2. Substitute S1 = −b and S2 = c: this becomes (−b)^2 − 2c = b^2 − 2c. So the sum of the squares is b^2 − 2c.

The key idea is to relate the sum of the squares of the roots to the sums of the roots and their pairwise products, using Vieta’s formulas for a monic quartic.

Let the roots be α, β, γ, δ. Define S1 = α+β+γ+δ and S2 = αβ + αγ + αδ + βγ + βδ + γδ. A monic quartic with these roots has the form x^4 − S1 x^3 + S2 x^2 − S3 x + S4, so the coefficients give S1 = −b and S2 = c.

The sum of squares can be written as α^2+β^2+γ^2+δ^2 = (α+β+γ+δ)^2 − 2(αβ+αγ+αδ+βγ+βδ+γδ) = S1^2 − 2S2.

Substitute S1 = −b and S2 = c: this becomes (−b)^2 − 2c = b^2 − 2c.

So the sum of the squares is b^2 − 2c.