Multiple choice

Determine the values of $p$ for which the quadratic equation $2x^2 + px + 8 = 0$ has real roots.

  1. $p < - 8$ or $p > 8, p \in R$
  2. $p \geq - 8$ and $ p \leq 8, p \in R$
  3. $p \leq - 8$  or $p \geq 8, p \in R$
  4. $p > - 8$ and $ p < 8, p \in R$
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C Correct answer
Explanation

For real roots, the discriminant D = b^2 - 4ac must be >= 0. Here, p^2 - 4(2)(8) >= 0, so p^2 - 64 >= 0. This implies p^2 >= 64, which means p <= -8 or p >= 8.

AI explanation

For the quadratic equation to have real roots, its discriminant must be greater than or equal to zero. The discriminant is b squared minus 4ac, so substituting a equals 2, b equals p, and c equals 8 gives p squared minus 64 is greater than or equal to 0. Solving this inequality yields p squared is greater than or equal to 64. Taking the square root of both sides gives the result that p is less than or equal to negative 8 or p is greater than or equal to 8.

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