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

LookingForAnswers Jan 7, 2021

#1**0 **

x^2-6x+k=0

Factor the equation into (x+a)(x+b)=0

We know that a+b=-6 and the maximum value of ab is 9

:D

AvenJohn Jan 7, 2021

#2**0 **

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

Rearrange as x^2 - 6x + k = 0

If this has at least one real solution, the discriminant must be ≥ 0

So

(-6)^2 - 4(1)(k) ≥ 0

36 - 4k ≥ 0

36 ≥ 4k

9 ≥ k

So .....the max value of k that produces at least one real solution is when k = 9

hihihi Jan 7, 2021