Multiple choice

If the roots of the equation $\lambda^{2}+8\lambda+\mu^{2}+6\mu=0$ are real, then $\mu$ lies between

  1. $-2$ and $8$
  2. $-3$ and $6$
  3. $-8$ and $2$
  4. $-6$ and $3$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For the roots of lambda^2 + 8lambda + (mu^2 + 6mu) = 0 to be real, the discriminant D must be >= 0. D = 8^2 - 4(mu^2 + 6mu) = 64 - 4mu^2 - 24mu >= 0. Dividing by -4: mu^2 + 6mu - 16 <= 0. Factors: (mu+8)(mu-2) <= 0. This holds for -8 <= mu <= 2.

AI explanation

For the roots of the quadratic equation in lambda to be real, the discriminant must be greater than or equal to zero, so 8^2 - 4(1)(mu^2 + 6mu) >= 0. Simplifying this inequality gives 64 - 4mu^2 - 24mu >= 0. Dividing by -4 and flipping the inequality sign yields mu^2 + 6mu - 16 <= 0. Factoring the quadratic gives (mu + 8)(mu - 2) <= 0, which means mu lies between -8 and 2.

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