Multiple choice

The equation $\displaystyle 9y^{2}(m+3)+6(m-3)y+(m+3)=0 $, where $m$ is real has real roots then

  1. $\displaystyle m< 0$
  2. $\displaystyle m> 0$
  3. $\displaystyle m\leq 0$
  4. $\displaystyle m\geq 0$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For real roots, D >= 0. D = [6(m-3)]^2 - 4(9(m+3))(m+3) = 36(m-3)^2 - 36(m+3)^2 = 36[(m-3)^2 - (m+3)^2] = 36[m^2-6m+9 - (m^2+6m+9)] = 36[-12m] = -432m. For D >= 0, -432m >= 0, so m <= 0.

AI explanation

For the quadratic equation to have real roots, its discriminant must be greater than or equal to zero, which gives [6(m-3)]^2 - 4[9y^2 does not apply, we use the coefficients directly as 6^2(m-3)^2 - 4(9(m+3))(m+3) >= 0. Simplifying this inequality yields 36(m-3)^2 - 36(m+3)^2 >= 0. Expanding both squares gives 36[(m^2 - 6m + 9) - (m^2 + 6m + 9)] >= 0, which reduces to -72m >= 0. Solving for m gives m <= 0.

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