if the cost, C, for manufacturing x units of a certain product is given by C=x^2-5x+70, find the number of units manufactured at a cost of $8620

- anonymous

- chestercat

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- richyw

well \(C(x)=x^2-5x+70\) and you know the cost is $8620 so \[x^2-5x+70=8620\]now solve for x. Do you know how to do that?

- anonymous

no, that is where I'm having the problem cause I don't understand algebra

- anonymous

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## More answers

- anonymous

quadratic formula, where a = 1, b= 5, c = -8550

- anonymous

b = -5 sorry

- richyw

alright well the first thing you usually want to do is write this in general form. To do this you subtract 8620 from both sides

- anonymous

I have no idea what you are talking about

- richyw

\[x^2-5x+70-8620=0\]\[x^2-5x-8850=0\]Now you want to find the roots to that. First you want to see if it can be factored, if it can then that's great and if not you need to use the quadratic formuala.

- richyw

i'm getting there!

- anonymous

ok

- richyw

the quadratic formula finds the roots of quadratic polynomials.
For a quadratic polynomial in general form, that is \(f(x)=ax^2+bx+c=0\), the quadratic formulas is \[\frac{-b\pm\sqrt{b^2-4ac}}{2a}\]

- richyw

so looking at the equation we have \(x^2-5x-8850\), you can pick out the parameters a,b and c.

- richyw

here the \(x^2\) term has nothing in front of it, so \(a=1\), the \(x\) term has a -5 in front of it so \(b=-5\), and \(c=8850\). Does that make sense?

- richyw

sorry \(c=-8850\)

- richyw

now plug those numbers into the the quadratic equation.

- anonymous

sorry I'm not responding....... i'm writing and studying this so I can understand it

- anonymous

I don't even know what the quadratic equation is

- richyw

yup MEMORIZE the quadratic formula! also do you understand what the \(\pm\) sign means?

- anonymous

no....... this is all new to me....... I didn't take algebra in high school

- richyw

alright well a quadratic EQUATION is a polynomial equation of the second degree. The initial problem you have is a quadratic equation. The quadratic FORMULA is the formula I gave you. You use the quadratic formula to find the roots of quadratic equations.

- richyw

This is the quadratic FORMULA.\[\frac{-b\pm\sqrt{b^2-4ac}}{2a}\]

- anonymous

ok

- richyw

the \(\pm\)is just used to save space, it means that this is actually two formulas...haha\[\frac{-b+\sqrt{b^2-4ac}}{2a}\]\[\frac{-b-\sqrt{b^2-4ac}}{2a}\]

- anonymous

I have been on this type of problem now for 2 hours

- richyw

but this equation is only useful if your equation is in general form. That was the first step that we did (when we subtracted 8620)

- anonymous

ok

- richyw

right so now just plug the numbers in \[\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{-(-5)\pm\sqrt{(-5)^2-4(1)(-8550)}}{2(1)}\]

- richyw

this gives \[x=\frac{5+\sqrt{25+4(8550)}}{2}\]\[x=\frac{5-\sqrt{25+4(8550)}}{2}\]

- richyw

For the context of the question though you can't have negative sales, you you only need to plug in the top one...

- richyw

sorry! how could I have forgotten this, if you were confused before, the quadratic equation gives you x... I'll put in an x= there for you :D. Anyways I don't have a calculator, so plug those two above in.\[x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\]

- anonymous

for the top I came up 34225

- richyw

nope that's not correct, you must have messed up on the calculator. Hand on i'll just check wolfram...

- anonymous

ok... I'm not good with math.....

- richyw

well we already finished the math, now it's just using the calculator :). But the two roots of this polynomial are x=-90 and x=95. Obviously the answer to the question is x=95, because you can't manufacture a negative number of products.
Just you know you can always verify this on wolfram alpha
http://www.wolframalpha.com/input/?i=x%5E2-5x%2B70%3D8620

- richyw

It might seem compplicated but really all you did was two steps
1)Put equation in general form
2)Find the roots (in this case we found the roots using the quadratic equation).

- anonymous

ty

- richyw

No worries, and good luck with your math studies! I only learned algebra after high school as well. It just takes a LOT of work! It is sure worth it though.

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