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

Guest Feb 27, 2022

#1**+1 **

Rearrange the equation, x^{2 }- 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^{2 }- 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.

Guest Feb 27, 2022

#2**+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

XxmathguyxX Feb 27, 2022