+0  
 
0
119
2
avatar

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

 Feb 27, 2022
 #1
avatar
+1

Rearrange the equation, x- 6x + k = 0 , then solve using the quadratic formula.
You will realise that the only possibility for x to be real is when the term under the square root (b- 4ac) must be greater than or equal to 0, so you do not get a negative result under the square root (which makes it imaginary).

You may read more on "discriminants" for more information.

 Feb 27, 2022
 #2
avatar+361 
+1

ok, so

6x - x^2 = k

everytime x is increased by 1, 6x is increased by 6

everytime x is increased by 1, x^2 is increased by 2x+1, since,

\((x+1)\times(x+1)\)

\(=x(x+1)+1(x+1)\)

\(=x^2+x+x+1\)

so we added 2x+1

 

I'll let you figure this out 

 

 

 

p.s also trial and error will work just as fine

 Feb 27, 2022

17 Online Users

avatar
avatar