For a cubic ax^3 + bx^2 + cx + d = 0, express Σ r_i^2 via coefficients.
Solution
Correct Answer: Option A
Let the roots be \(r_1,r_2,r_3\) of \(ax^3+bx^2+cx+d=0\). By Viète's relations,
\[
r_1+r_2+r_3=-\frac{b}{a},\qquad
r_1r_2+r_2r_3+r_3r_1=\frac{c}{a},\qquad
r_1r_2r_3=-\frac{d}{a}.
\]
Use the identity
\[
r_1^2+r_2^2+r_3^2=(r_1+r_2+r_3)^2-2(r_1r_2+r_2r_3+r_3r_1).
\]
Substitute the Viète expressions:
\[
r_1^2+r_2^2+r_3^2=\Big(-\frac{b}{a}\Big)^2-2\Big(\frac{c}{a}\Big).
\]
Thus the correct choice is Option 1: \(\displaystyle\big(-\frac{b}{a}\big)^2-2\frac{c}{a}.\)
\[
r_1+r_2+r_3=-\frac{b}{a},\qquad
r_1r_2+r_2r_3+r_3r_1=\frac{c}{a},\qquad
r_1r_2r_3=-\frac{d}{a}.
\]
Use the identity
\[
r_1^2+r_2^2+r_3^2=(r_1+r_2+r_3)^2-2(r_1r_2+r_2r_3+r_3r_1).
\]
Substitute the Viète expressions:
\[
r_1^2+r_2^2+r_3^2=\Big(-\frac{b}{a}\Big)^2-2\Big(\frac{c}{a}\Big).
\]
Thus the correct choice is Option 1: \(\displaystyle\big(-\frac{b}{a}\big)^2-2\frac{c}{a}.\)
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