Multiple choice

The possible values of p for which the equation $\displaystyle { x }^{ 2 }+px+64=0$ and $\displaystyle { x }^{ 2 }-8x+p=0$ will both have real roots is

  1. $16$
  2. $8$
  3. $4$
  4. $12$
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A Correct answer
Explanation

For x^2 + px + 64 = 0, D = p^2 - 256 >= 0 implies |p| >= 16. For x^2 - 8x + p = 0, D = 64 - 4p >= 0 implies p <= 16. Both conditions are satisfied only when p = 16 or p = -16. Since 16 is an option, it is the correct value.

AI explanation

For both quadratic equations to have real roots, their discriminants must be greater than or equal to zero. For x squared plus px plus 64 equals 0, p squared minus 256 is greater than or equal to 0, meaning p squared is greater than or equal to 256. For x squared minus 8x plus p equals 0, 64 minus 4p is greater than or equal to 0, meaning p is less than or equal to 16. The only value among the options that satisfies both conditions, meaning p must be plus or minus 16 and also less than or equal to 16, is 16.

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