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Leya [2.2K]
3 years ago
7

What is the largest value of $k$ such that the equation $6x - x^2 = k$ has at least one real solution?

Mathematics
2 answers:
podryga [215]3 years ago
5 0

Step-by-step explanation:

6x - x² = k

=> x² - 6x + k = 0,

For there to be at least 1 real solution,

The discriminant b² - 4ac must be at least 0.

We have (-6)² - 4(1)(k) >= 0.

Therefore 36 - 4k >= 0, k <= 9.

Hence the largest value of k is 9.

amid [387]3 years ago
3 0

Answer:

9

Step-by-step explanation:

Rearranging, this is the equation x^2 - 6x  = -k. Completing the square, we add 9 to both sides, and x^2 - 6x + 9 = 9 - k. Factoring the left side, we have (x - 3)^2 = 9 - k. In order for this equation to have at least one real solution, the right-hand side must be nonnegative. Thus, we have 9 - k \geq 0 and k \leq 9. Thus, the largest possible value of k is 9.

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