Algebra

The Discriminant of the Quadratic Equation in the Standard Form \(ax^2+bx+c=0\)  is equal to \(b^2-4ac\)

In order to an Quadratic Equation to have 2 real Solutions , the Discrimnant should be greater than 0 ( \(​​b^2-4ac>0\) )

In order to an Quadratic Equation to have 1 real Solutions , the Discrimnant should be equal 0 ( \(b^2-4ac=0\) )

In order to an Quadratic Equation to have at least 1 real Solutions , the Discrimnant should be greater than or equal 0 ( \(b^2-4ac\ge0\) )

So for our Question,  we have the Equation \(6x-x^2=k\)

Rearrange it to put in the Standard Form , \(-x^2+6x-k=0\)

Then \(a=-1, b=6 , c=-k\)

Then the Discriminant \(b^2-4ac=6^2-4(-1)(-k)=36-4k\)

For at least one Real Solution , \(b^2-4ac\ge0\)

\(36-4k\ge0\)\(\)

Solve for k , by adding 4k for both sides

\(36\ge4k \)

Dividing both sides by 4

\(\frac{36}{4}\ge\frac{4k}{4}\)

Then \(k\le9\)

So the Largest Value of k that makes the Equation has at least one Solution is \(9\)