Question

Solve A=2(Π)r2+2(Π)rh for positive r." Hint: Use Quadratic Formula.
Help? Medal will be given.

Answer 1

Do you know the formula for solving any quadratic equation?

Answer 2

\(\bf A=2\pi r^2+2\pi rh\implies A=2\pi (r^2+rh)\implies 0=2\pi (r^2+rh)
\\ \quad \\
0=r^2+rh\implies 0={\color{blue}{ 1}}r^2+{\color{red}{ h}}r{\color{green}{ +0}}
\\ \quad \\
\textit{quadratic formula}\\
\qquad
x= \cfrac{ - {\color{red}{ b}} \pm \sqrt { {\color{red}{ b}}^2 -4{\color{blue}{ a}}{\color{green}{ c}}}}{2{\color{blue}{ a}}}\)

Answer 3

Okay, I understand that part. But once I solved it and got \[x=(-h+h)/2\]
Is that what r equals?? And in that case, the numerator always cancels to 0, correct? So r=0??

Answer 4

I think that is making an implicit assumption that A = 0, which is not generally true.

Instead, we should treat A as another constant.
\( A = 2 \pi r^2 + 2 \pi r h \)
To get the general form of a quadratic, we subtract both sides by A.
\( 0 = \color{red}{2 \pi } \ r^2 + \color{red}{2 \pi h} \ r \color{red}{- A} \)
This is now a quadratic equation in r. Try not to be intimidated by the level of abstraction here: just work it down like any quadratic using the formula you already used, using a = \(2\pi\), b = \(2 \pi h\), and c = \(-A\).

Answer 5

@AccessDenied how would you simplify =/- rt of [(2*pi*h)^2 + 8*pi*A} over 2*pi