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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.2012-07-16

4 Answers 4

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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

$\dfrac{Ah}{2\pi}=r^2+r$

then "taking square roots" produces

$\sqrt{\dfrac{Ah}{2\pi}}=\sqrt{r^2+r}$

You then seem to assert

$\sqrt{r^2+r}=2r$

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

$\sqrt{2}=2$

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

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

COMPLETING THE SQUARE:

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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    @AustinBroussard Look again. Your $x$ is $r$, not $rh$.2012-07-16
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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:

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

$\dfrac{Ah}{2\pi}=r^2+r$

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:

$0=r^2+r-\dfrac{Ah}{2\pi}$

And then applying our friend the quadratic formula.