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 \(9\)
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