Multiple choice

If $\alpha$ and $\beta$ are the roots of the equation $3x^2+2x+1=0$, then the equation whose roots are $\alpha+ \beta^{-1}$ and $\beta+\alpha^{-1}$ is

  1. $3x^2+8x+16= 0$
  2. $3x^2-8x-16 = 0$
  3. $3x^2+8x-16 = 0 $
  4. $x^2+8x+16 = 0 $
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A Correct answer
Explanation

Roots are a, b. a+b = -2/3, ab = 1/3. New roots are a + 1/b = (ab+1)/b and b + 1/a = (ab+1)/a. Sum = (ab+1)(a+b)/ab = (4/3 * -2/3) / (1/3) = -8/3. Product = (ab+1)^2 / ab = (4/3)^2 / (1/3) = (16/9) * 3 = 16/3. Equation: x^2 - (sum)x + product = 0 -> x^2 + 8/3x + 16/3 = 0 -> 3x^2 + 8x + 16 = 0.

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