Multiple choice

For what value of a, the roots of the equation $2x^2+ 6x + a = 0$, satisfy the conditions $\left ( \dfrac{\alpha }{\beta } \right )+\left ( \dfrac {\beta}{\alpha } \right ) < 2 $ (where $\beta,\alpha $ are the roots of equation).

  1. $a > 0$
  2. $-1 < a < 0$
  3. $-1 < a < 1$

  4. None of these
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

(a/b + b/a) < 2 => (a^2+b^2)/ab < 2. Since (a-b)^2 >= 0, a^2+b^2 >= 2ab. For the sum to be less than 2, the roots must be complex or have different signs. Discriminant D = 36 - 8a. For real roots, a <= 4.5. The condition (a/b + b/a) < 2 is impossible for real roots of the same sign. Thus, no real 'a' satisfies this.

AI explanation

The condition (alpha divided by beta) plus (beta divided by alpha) less than 2 simplifies to the fraction (alpha plus beta) squared divided by (alpha times beta) minus 2 being less than 2, which means (alpha plus beta) squared divided by (alpha times beta) is less than 4. Using sum and product of roots, this becomes (-6 divided by 2) squared divided by (a divided by 2) is less than 4, so 9 divided by a is less than 4. For positive a, this requires a greater than 2.25, and for negative a, the inequality is always true but the discriminant 36 minus 8a must be positive, meaning there is no single comprehensive range provided in the options. The result is None of these.

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