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Solve for $+r$

$$A=2\pi r^2+2\pi rh$$ Since $2\pi$ is common on both sides of the $+$ so I will take it out
$$A=2\pi (r^2+rh)$$ Now, divide both sides by $2\pi$ $$\dfrac{A}{2\pi}=r^2+rh$$ Then, we can divide by the $h$ $$\dfrac{Ah}{2\pi}=r^2+r$$ Then; $$0=r^2+r-\dfrac{Ah}{2\pi}$$ Quadratic formula $=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ $a=1,b=1,c=?$ What are the values of $a,b,c$

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I'm ultimately trying to isolate the variable $r$ in a positive form. – Austin Broussard Jul 16 '12 at 3:45
up vote 3 down vote accepted

There is a series of mistakes in your solution. I'll mark them one by one:

You correctly arrive to

$$\frac A {2\pi}=r^2+rh$$

But "dividing by $h$" produces

$$\frac 1 h \frac A {2\pi}=\frac 1 h\left(r^2+rh\right)$$

$$\frac 1 h \frac A {2\pi}=\frac {r^2} h+\frac{rh}h$$

$$\frac 1 h \frac A {2\pi}=\frac {r^2} h+r$$

So that step is wrong.

Similarily, if you have


then "taking square roots" produces


You then seem to assert


Let's check if it is indeed true for, say $r=1$, which gives


which is manifestly false. So there is something awry there, too.

The best thing you can do is check wether each step is correct. To solve for $r$, since

$$\frac A {2\pi}=r^2+rh$$

is a quadratic we need to make a very old trick, which is called completing the square:

$$\eqalign{ & \frac{A}{{2\pi }} = {r^2} + rh \cr & \frac{A}{{2\pi }} = {r^2} + 2r\frac{h}{2} \cr & \frac{A}{{2\pi }} = \underbrace {{r^2} + 2r\frac{h}{2} + {{\left( {\frac{h}{2}} \right)}^2}}_{{\text{This is a perfect square!}}} - {\left( {\frac{h}{2}} \right)^2} \cr & \frac{A}{{2\pi }} = {\left( {r + \frac{h}{2}} \right)^2} - {\left( {\frac{h}{2}} \right)^2} \cr & \frac{A}{{2\pi }} + {\left( {\frac{h}{2}} \right)^2} = {\left( {r + \frac{h}{2}} \right)^2} \cr & \sqrt {\frac{A}{{2\pi }} + {{\left( {\frac{h}{2}} \right)}^2}} = {{{\left( {r + \frac{h}{2}} \right)}^2}} \cr & \pm \sqrt {\frac{A}{{2\pi }} + {{\left( {\frac{h}{2}} \right)}^2}} = {r + \frac{h}{2}} \cr} $$

Note in the last steps we take the square root. We then have to think about both the positive and negative root. So you final solution is

$$r = - \sqrt {\frac{A}{{2\pi }} + {{\left( {\frac{h}{2}} \right)}^2}} - \frac{h}{2}{\text{ or }}r = \sqrt {\frac{A}{{2\pi }} + {{\left( {\frac{h}{2}} \right)}^2}} - \frac{h}{2}$$


Say we have a quadratic $$0=ax^2+bx+c$$

"Completing the square" consist of writing it in the form

$$0 = A{\left( {x + h} \right)^2} + C$$

We can accomplish this with some "trickery"

$$\eqalign{ & 0 = a{x^2} + bx + c \cr & 0 = 4{a^2}{x^2} + 4abx + 4ac{\text{ ; multiply by }}4a \cr & 0 = {\left( {2ax} \right)^2} + 2 \cdot \left( {2ax} \right) \cdot b + 4ac{\text{ ; cleverly rearrange the eqn}}{\text{.}} \cr & {b^2} = {\left( {2ax} \right)^2} + 2 \cdot \left( {2ax} \right) \cdot b + {b^2} + 4ac{\text{ ; add }}{b^2} \cr & {b^2} = \underbrace {{{\left( {2ax} \right)}^2} + 2 \cdot \left( {2ax} \right) \cdot b + {b^2}}_{{\text{This is a perfect square!}}} + 4ac \cr & {b^2} = {\left( {2ax + b} \right)^2} + 4ac \cr & {b^2} - 4ac = {\left( {2ax + b} \right)^2} \cr & \sqrt {{b^2} - 4ac} = 2ax + b \cr & - b + \sqrt {{b^2} - 4ac} = 2ax \cr & \frac{{ - b + \sqrt {{b^2} - 4ac} }}{{2a}} = x \cr} $$

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I have forgotten exactly how completing the square works. I know it is a way of solving a quadratic equation. Do you think you could shed some light on that using $0=ax^2+bx+c$. – Austin Broussard Jul 16 '12 at 3:50
@AustinBroussard You could set $$0 = {r^2} + rh - \frac{A}{{2\pi }}$$ but you'll arrive to the same result and it will be a little more cumbersome, IMO. – Pedro Tamaroff Jul 16 '12 at 3:57
Right, but could you sort of "teach" me completing the square again. It would help me grasp the concept of it for future problems. – Austin Broussard Jul 16 '12 at 3:59
@AustinBroussard Oh, sure. – Pedro Tamaroff Jul 16 '12 at 3:59
Thanks for the extra help. – Austin Broussard Jul 16 '12 at 4:02

No! $\sqrt{r^2 + r} \neq 2r.$

What you have is a polynomial equation $r^2 + r - \dfrac{Ah}{\pi} = 0.$

Since this is a homework, I will only give a hint: the two roots $r_1, r_2$ of quadratic $ax^2 + bx + c = 0$ are $$ r_{1,2} = \frac{-b \pm \sqrt{b^2 -4ac}}{2a}.$$

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Hint: Try subtracting $2\pi r^2 + 2\pi rh$ from both sides of the original and completing the square. I can expand on this if you need me to.

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You've made two errors that I can spot first from here to here:



Dividing by h doesn't work like that.

Your last square root is incorrect. You should be looking at the problem in the form of:


And then applying our friend the quadratic formula.

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