Multiple choice

If $\alpha, \beta, \gamma $ are the roots of the equation $2x^3-3x^2+6x+1=0$, then $\alpha^2 +\beta^2+ \gamma^2$ is equal to

  1. $-\dfrac{15}{4}$
  2. $\dfrac{15}{4}$
  3. $\dfrac{9}{4}$
  4. $4$
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A Correct answer
Explanation

For a cubic equation ax^3 + bx^2 + cx + d = 0, the sum of roots alpha + beta + gamma = -b/a = 3/2. The sum of roots taken two at a time alpha*beta + beta*gamma + gamma*alpha = c/a = 6/2 = 3. We use the identity alpha^2 + beta^2 + gamma^2 = (alpha + beta + gamma)^2 - 2(alpha*beta + beta*gamma + gamma*alpha). Substituting values: (3/2)^2 - 2(3) = 9/4 - 6 = 9/4 - 24/4 = -15/4.

AI explanation

By Vieta's formulas for the cubic equation 2x^3 - 3x^2 + 6x + 1 = 0, the sum of the roots (alpha + beta + gamma) equals -(-3)/2 = 3/2. We use the algebraic identity alpha^2 + beta^2 + gamma^2 = (alpha + beta + gamma)^2 - 2(alpha*beta + beta*gamma + gamma*alpha). The sum of the product of the roots taken two at a time is 6/2 = 3, so substituting these values gives (3/2)^2 - 2(3) = 9/4 - 6 = -15/4.

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