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# Algebra

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

Nov 8, 2022

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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 $$9$$

Nov 8, 2022