Express r1^2 + r2^2 for a quadratic ax^2 + bx + c = 0 in terms of coefficients.
Solution
Correct Answer: Option A
Start with Viète's relations for the quadratic ax^2 + bx + c = 0:
r1 + r2 = −b/a, r1 r2 = c/a.
Compute
\[
r_1^2 + r_2^2 = (r_1 + r_2)^2 - 2r_1r_2
= \Big(\!-\frac{b}{a}\!\Big)^2 - 2\cdot\frac{c}{a}
= \frac{b^2}{a^2} - \frac{2c}{a}
= \frac{b^2 - 2ac}{a^2}.
\]
Thus the correct choice is Option 1: \(\displaystyle \Big(-\frac{b}{a}\Big)^2 - 2\frac{c}{a}.\) (Option 4, which has a plus sign, is incorrect.)
r1 + r2 = −b/a, r1 r2 = c/a.
Compute
\[
r_1^2 + r_2^2 = (r_1 + r_2)^2 - 2r_1r_2
= \Big(\!-\frac{b}{a}\!\Big)^2 - 2\cdot\frac{c}{a}
= \frac{b^2}{a^2} - \frac{2c}{a}
= \frac{b^2 - 2ac}{a^2}.
\]
Thus the correct choice is Option 1: \(\displaystyle \Big(-\frac{b}{a}\Big)^2 - 2\frac{c}{a}.\) (Option 4, which has a plus sign, is incorrect.)
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