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Find all values of k for which the equation has two complex (non-real) solutions. 7z²+(k–1)=– 9z

User Cbrauchli
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Final answer:

The values of k for which the equation 7z² + (k–1) = – 9z has two complex solutions are found by setting the discriminant b² - 4ac of the quadratic formula negative and solving for k, which yields k > 109/28.

Step-by-step explanation:

The question is asking to find the values of k for which the quadratic equation 7z² + (k–1) = – 9z has two complex solutions. To determine when a quadratic equation has complex solutions, we look at the discriminant of the quadratic formula, b² - 4ac, which must be negative for complex solutions.

For the equation 7z² + (k–1) = – 9z, we rewrite it in the form az² + bz + c = 0 to give 7z² + 9z + (k – 1) = 0. Here, a=7, b=9, and c=k – 1. The discriminant is thus b² - 4ac = 9² - 4(7)(k - 1).

To find when this discriminant is negative, we solve the inequality 9² - 4(7)(k - 1) < 0. Simplifying, we have 81 - 28k + 28 < 0 which simplifies to 109 - 28k < 0. Solving for k gives us k > 109/28. Therefore, the quadratic equation has two complex solutions when k is greater than 109/28.

User Hiws
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