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-x² + 42x + j = 0

In the given equation, j is a constant. The equation has no real solutions. If j > k, what is the maximum possible integer value of k?

User Luke Tan
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1 Answer

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Final answer:

The maximum possible integer value of k for the equation with no real solutions is -442, considering that the discriminant must be negative with j being greater than k.

Step-by-step explanation:

To find the maximum possible integer value of k when given an equation with no real solutions, we need to look at the discriminant of the quadratic equation.

The equation provided is -x² + 42x + j = 0. For a quadratic equation ax² + bx + c = 0, the discriminant is given by b² - 4ac.

For the equation to have no real solutions, the discriminant must be negative, so we should have 42² - 4(-1)j < 0.

After calculating, we get 1764 + 4j < 0. If we solve for j, we find that j < -441. Since we know that j > k, the maximum possible integer value for k would be just less than -441, which is -442.

User Nate Zaugg
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