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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?

 Jan 7, 2021
 #1
avatar+218 
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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

 Jan 7, 2021
 #2
avatar+303 
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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?

 

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

 Jan 7, 2021

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