Multiple choice

lf the roots of the equation $x^{2}+2ax+b=0$ are real and distinct and they differ by at most $2m$, then $b$ lies in the interval

  1. $(\mathrm{a}^{2}-\mathrm{m}^{2}, \mathrm{a}^{2})$
  2. $[\mathrm{a}^{2}-\mathrm{m}^{2}, \mathrm{a}^{2})$
  3. $( \mathrm{a}^{2} -\mathrm{m}^{2}, \infty)$
  4. $[\mathrm{a}^{2}-\mathrm{m}^{2},\infty ]$
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B Correct answer
Explanation

Roots are x = -a +/- sqrt(a^2 - b). Difference = 2 * sqrt(a^2 - b). Given 2 * sqrt(a^2 - b) <= 2m, so sqrt(a^2 - b) <= m. Squaring gives a^2 - b <= m^2, so b >= a^2 - m^2. Also, for real distinct roots, a^2 - b > 0, so b < a^2.

AI explanation

Let the roots of the quadratic equation be alpha and beta. The difference of the roots is given by the formula |alpha - beta| = sqrt((2a)^2 - 4b) = 2sqrt(a^2 - b). Since the roots differ by at most 2m, we have 2sqrt(a^2 - b) <= 2m, which simplifies to sqrt(a^2 - b) <= m. Squaring both sides gives a^2 - b <= m^2, meaning b >= a^2 - m^2. Furthermore, because the roots are real and distinct, the discriminant must be positive, so 4a^2 - 4b > 0, which gives b < a^2. Combining these inequalities, b lies in the interval [a^2 - m^2, a^2).

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