Multiple choice

If $ \alpha, \beta $ are roots of the equation $ 2x^{2}+6x+b=0&gt;(b<0)$, then $ \dfrac{\alpha}{\beta}+\dfrac{\beta}{\alpha} $ is less than

  1. $2$
  2. $-2$
  3. $18$
  4. $0$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For 2x^2 + 6x + b = 0, the sum of roots alpha + beta = -6/2 = -3 and product alpha*beta = b/2. The expression alpha/beta + beta/alpha = (alpha^2 + beta^2) / (alpha*beta) = ((alpha + beta)^2 - 2*alpha*beta) / (alpha*beta) = (9 - b) / (b/2) = 18/b - 2. Since b < 0, 18/b is negative, so 18/b - 2 < -2.

AI explanation

Combine the fractions in the given expression to get (alpha^2 + beta^2)/(alpha * beta). Rewrite the numerator using the identity (alpha + beta)^2 - 2(alpha * beta), so the expression becomes ((alpha + beta)^2)/(alpha * beta) - 2. From the equation 2x^2 + 6x + b = 0, Vieta's formulas give the sum of the roots as alpha + beta = -3 and the product as alpha * beta = b/2. Substituting these values gives ((-3)^2)/(b/2) - 2, which simplifies to 18/b - 2. Since b is negative, the term 18/b is a negative number, making the total expression (a negative number) - 2, which is strictly less than -2.

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