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 - 4*1*64 = k² - 256. This must be greater than or equal to 0. So, k² - 256 = 0. Solving this inequality, we get k = 16 or k = 0. Solving this inequality, we get k

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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 - 4*1*64 = 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 - 4*1*k = 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.

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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 - 4*1*64 = 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 - 4*1*k = 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.

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

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Find the positive value of k for which the equations:

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