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Was wondering whether you might be able to take a look over my working and see if I'm doing this right! I was using the box method before, but have tried to do this one this way!

Find the break-even points in the case where total cost function $TC=7+2x+x^2$ and total revenue function is $TR=10x$.

Therefore, we need to solve: $$TC = TR$$

$$7+2x+x^2=10x\\ \therefore x^2+2x+7-10x=0\\ \therefore x^2-8x+7 = 0 $$ Now having the equation in the quadratic form $ax^2+bx+c$, we can solve using the quadratic formula: $$x = \frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ $$ Plugging in the values $a=1$, $b=-8$ and $c=7$, we get: $$\begin{align} x & = \frac{-(-8)\pm \sqrt{(-8)^2-(4\cdot1\cdot7)}}{2\cdot1}\\\\ & = \frac{8\pm \sqrt{64-28}}{2}\\\\ & = \frac{8\pm \sqrt{36}}{2}\\\\ & = \frac{8\pm 6}{2}\\ \end{align} $$ Therefore $x$ can equal: $$\begin{align}\\ x & =\frac{8+6}{2}\\\\ & = \frac{14}{2}\\\\ & = 7 \end{align} $$ OR: $$\begin{align}\\ x & =\frac{8-6}{2}\\\\ & = \frac{2}{2}\\\\ & = 1 \end{align} $$ Therefore, the break-even points could equal either $7$ or $1$.

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  • $\begingroup$ Looks good! ${}{}{}{}$ $\endgroup$ – Namaste Nov 18 '16 at 21:22
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Yes, your follow through is correct. Although I would like to point out one thing: You didn't have to use the quadratic formula to find the roots of the polynomial.

Whenever you get a polynomial, factoring usually is the quickest way to find the roots.

For your polynomial, $x^2-8x+7=0$, we see that we need two numbers to have a product of $7$ and sum to $8$. $1,7$ fit quite nicely. Thus, your quadratic can be factored into$$x^2-8x+7=(x-1)(x-7)=0\tag1$$ From which we see $x=1,7$ the two solutions.

Personally, I think it's quicker to factor than use the quadratic formula. Although the quadratic formula does have the nice feature of working on $any$ polynomial.

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