SOLUTION: I am not sure if this is a unit conversion sorry for any confusion. 75. If the cost, C(x), for manufacturing x units of a certain product is given by {{{ C(x)=x^2-15x+50 }}}, fi

Question 72609: I am not sure if this is a unit conversion sorry for any confusion.
75. If the cost, C(x), for manufacturing x units of a certain product is given by , find the number of units manufactured at a cost of $9500.
I have tried factoring this by first factoring the equation without the 9500 even inputed. I have tried substituting the 9500 in to the equation in a couple of different ways. The book has given me the answer and if I put the answer into the equation I see that it works but I have no idea how the number was gotten in the first place. There is no explanation in the book please help. Thank you
Answer by bucky(2189) (Show Source): You can put this solution on YOUR website!

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Substitute 9500 for C(x):
.

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Subtract 9500 from both sides and transpose to get:
.

.
This will factor into:
.
.
.
and you set each of the two factors equal to zero to find that either x = 105 or x = -90 are
the solutions to this equation. And since it makes no sense to have a minus number of
units manufactured, the correct answer is x = 105 units.
.
But a more general solution exists. It is the quadratic formula and it is based on a method
called completing the squares. You can just know that it exists so you can look it up if you
need to use it, or you can (as most people that work math generally do) just memorize it.
This formula says that if you have a quadratic equation of the form:
.

.
the answers for x are given by:
.

.
By comparing with the equation we have you
can see that a = 1, b = -15, and c = -9450.
.
Substitute these numbers into the equation to solve for x and you get:
.

.
which simplifies to:
.

.
So the two answers for x are:
.

.
and
.

.
These are the same answers that we got by factoring, so we probably haven't made a mistake.
.
Notice that the quadratic formula is a general method of use. It will work on quadratic
equations when, as in this case, it is hard to find the factors or the numbers are so
big that it is difficult to do a graph with enough accuracy. It is well worth memorizing the
formula if you have lots of difficult quadratics to work on.
.
Hope this discussion helps to fill in the gaps left by the book you are using.

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