Multiple choice

If $\alpha,\beta$ are the roots of $x^{2}+6x+9=0$, then the equation whose roots are $\dfrac{1}{\alpha}, \dfrac{1}{\beta}$ is

  1. $2x^{2}+3x+18=0$
  2. $x^{2}+6x-9=0$
  3. $9x^{2}+6x+1=0$
  4. $9x^{2}-6x+1=0$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For the equation x^2 + 6x + 9 = 0, the roots are alpha = -3 and beta = -3. The new roots are 1/alpha = -1/3 and 1/beta = -1/3, so the equation is (x + 1/3)(x + 1/3) = 0, which is x^2 + (2/3)x + 1/9 = 0, or 9x^2 + 6x + 1 = 0.

AI explanation

For a quadratic equation ax^2 + bx + c = 0 with roots alpha and beta, the equation whose roots are 1/alpha and 1/beta is obtained by replacing x with 1/x and clearing the denominator, giving cx^2 + bx + a = 0. Applying this method to the given equation x^2 + 6x + 9 = 0, we substitute the coefficients to get 9x^2 + 6x + 1 = 0.

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