Find the positive value of k for which the equations: x2 + kx + 64 = 0 and x2 – 8x+ k = 0 will have real roots:(a) 12(b) 16(c) 18(d) 22

For a quadratic equation ax² + bx + c = 0 to have real roots, the discriminant (b² - 4ac) must be greater than or equal to 0.

1. For the first equation x² + kx + 64 = 0, the discriminant is (k)² - 4164 = k² - 256. This must be greater than or equal to 0. So, k² - 256 >= 0. Solving this inequality, we get k >= 16 or k <= -16.

2. For the second equation x² - 8x + k = 0, the discriminant is (-8)² - 41k = 64 - 4k. This must be greater than or equal to 0. So, 64 - 4k >= 0. Solving this inequality, we get k <= 16.

Combining the results from the two equations, we find that the only positive value of k that satisfies both conditions is k = 16. So, the answer is (b) 16.