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

User Ngoa
by
3.9k points

2 Answers

8 votes

Answer:

9

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.

User Jjmirks
by
4.9k points
6 votes

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.

User Salisha
by
5.0k points