Multiple choice

Find the value of $k$ for which the equation ${x}^{2}+kx+81=0$ and ${x}^{2}-6\sqrt {2}x+k=0$ has both real roots, $k> 0$

  1. $k=9$
  2. $k=18$
  3. $k=3\sqrt {2}$

  4. No such value
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B Correct answer
Explanation

For x^2 + kx + 81 = 0 to have real roots, the discriminant k^2 - 4(81) >= 0, so k^2 >= 324, implying k >= 18 (since k > 0). For x^2 - 6sqrt(2)x + k = 0 to have real roots, the discriminant (6sqrt(2))^2 - 4k >= 0, so 72 - 4k >= 0, implying k <= 18. Combining k >= 18 and k <= 18, the only solution is k = 18.

AI explanation

For the first equation x^2 + kx + 81 = 0 to have real roots, the discriminant must be non-negative, giving k^2 - 4(1)(81) >= 0, which means k^2 >= 324. Since k > 0, this requires k >= 18. For the second equation x^2 - 6*sqrt(2)x + k = 0 to have real roots, its discriminant must also be non-negative, giving 72 - 4k >= 0, which means k <= 18. The only value that satisfies both conditions is k = 18.

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