SOLUTION: Publishing costs. The cost, in dollars, of publishing x books is C(x)=40,000+20x+0.0001x^2. How many books can be published for 250,000? Do I set 40,000+20x+0.0001x^2=250,000 th

Question 30763: Publishing costs. The cost, in dollars, of publishing x books is C(x)=40,000+20x+0.0001x^2. How many books can be published for 250,000?
Do I set 40,000+20x+0.0001x^2=250,000 then subtract the 250,000 so the equation =0 and then use the quadratic formula to solve? Could someone show me the correct equation for this problem?
Thank you!
Answer by mbarugel(146) (Show Source): You can put this solution on YOUR website!
Yes, you got it right. The original equation is:

Subtracting 250,000, we get:

And then we can solve this using the quadratic formula:

Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)

Quadratic equation (in our case ) has the following solutons:

For these solutions to exist, the discriminant should not be a negative number.

First, we need to compute the discriminant : .

Discriminant d=484 is greater than zero. That means that there are two solutions: .

Quadratic expression can be factored:

Again, the answer is: 10000, -210000. Here's your graph:

As in any quadratic equation, we get two solutions. However, we can disregard the negative one, since it doesn't make sense in this case. The answer is that 10,000 books can be published.
I hope this helps!
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