+20 What is the largest value of $k$ such that the equation $6x - x^2 = k$ has at least one real solution?
+237 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
+285 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
