What is the largest value of k such that the equation 6x - x^2 = k has at least one real solution?
Rearrange the equation, x2 - 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 (b2 - 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.
+364 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
