Multiple choice

If the roots of the equation $x^{2}-ax+b=0$ are real and differ by a quantity which is less than $c(c > 0)$, then $b$ lies between

  1. $\dfrac{a^{2}-c^{2}}{4}$ and $\dfrac{a^{2}}{4}$
  2. $\dfrac{a^{2}+c^{2}}{4}$ and $\dfrac{a^{2}}{4}$
  3. $\dfrac{a^{2}-c^{2}}{2}$ and $\dfrac{a^{2}}{4}$
  4. $None\ of\ these$
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A Correct answer
Explanation

Let roots be r1, r2. r1+r2 = a, r1*r2 = b. |r1-r2| < c => (r1-r2)^2 < c^2 => (r1+r2)^2 - 4r1r2 < c^2 => a^2 - 4b < c^2 => 4b > a^2 - c^2 => b > (a^2 - c^2)/4. Also, for real roots, D >= 0 => a^2 - 4b >= 0 => b <= a^2/4.

AI explanation

Let the real roots of the quadratic equation be p and q, having a sum of p+q = a and a product of pq = b. The difference between the roots is given by the formula |p - q| = sqrt((p+q)^2 - 4pq) = sqrt(a^2 - 4b). Because this difference must be less than c, we write the inequality sqrt(a^2 - 4b) < c. Squaring both positive quantities yields a^2 - 4b < c^2, which simplifies to 4b > a^2 - c^2, or b > (a^2 - c^2)/4. For the roots to be real in the first place, the discriminant must be non-negative, meaning a^2 - 4b >= 0, or b <= a^2/4. Therefore, b must lie between (a^2 - c^2)/4 and a^2/4.

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