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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

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

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Other answers:

quadratic formula, where a = 1, b= 5, c = -8550
b = -5 sorry
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
I have no idea what you are talking about
\[x^2-5x+70-8620=0\]\[x^2-5x-8850=0\]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.
i'm getting there!
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 \[\frac{-b\pm\sqrt{b^2-4ac}}{2a}\]
so looking at the equation we have \(x^2-5x-8850\), you can pick out the parameters a,b and c.
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?
sorry \(c=-8850\)
now plug those numbers into the the quadratic equation.
sorry I'm not responding....... i'm writing and studying this so I can understand it
I don't even know what the quadratic equation is
yup MEMORIZE the quadratic formula! also do you understand what the \(\pm\) sign means?
no....... this is all new to me....... I didn't take algebra in high school
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.
This is the quadratic FORMULA.\[\frac{-b\pm\sqrt{b^2-4ac}}{2a}\]
the \(\pm\)is just used to save space, it means that this is actually two formulas...haha\[\frac{-b+\sqrt{b^2-4ac}}{2a}\]\[\frac{-b-\sqrt{b^2-4ac}}{2a}\]
I have been on this type of problem now for 2 hours
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)
right so now just plug the numbers in \[\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{-(-5)\pm\sqrt{(-5)^2-4(1)(-8550)}}{2(1)}\]
this gives \[x=\frac{5+\sqrt{25+4(8550)}}{2}\]\[x=\frac{5-\sqrt{25+4(8550)}}{2}\]
For the context of the question though you can't have negative sales, you you only need to plug in the top one...
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.\[x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\]
for the top I came up 34225
nope that's not correct, you must have messed up on the calculator. Hand on i'll just check wolfram...
ok... I'm not good with math.....
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
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).
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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