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How do you solve #22x-x^2=96#?

Answer 1

Nathan Axel

# x = 16 or 6#

#22x -x^2 =96#

#-x^2+22x-96=0#

#x^2 -22x +96=0#

#(x-16)(x-6)=0#

Hence, #x= 16 or 6#

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Answer 2

Julia Adams

#x=6 and 16#

So first we want to get the #x^2# to be positive. To do this we have to multiply both sides by a #(-1)#. #(-1)[22x-x^2=96]#

Giving us #-22x+x^2=-96#

Now we can get all numbers to one side of the equation making it equal to #0#. #x^2-22x+96=0#

Now we can factor. Since #-16*-6=96# We can now write #x^2-22x+96#

As #(x-16)(x-6)=0#

Setting both to #0# gives us #x-16=0# and #x-6=0#

Which gives us #x=16# and #x=6#

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Answer 3

Genesis Allen

To solve the equation 22x - x^2 = 96, follow these steps:

Rearrange the equation to set it equal to zero: x^2 - 22x + 96 = 0.
Factor the quadratic equation: (x - 12)(x - 10) = 0.
Set each factor equal to zero and solve for x: x - 12 = 0 or x - 10 = 0.
Solve for x in each equation: x = 12 or x = 10.

So, the solutions to the equation are x = 12 or x = 10.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.