Question

Given $A = 2\left( \pi \right){r^2} + 2\left( \pi \right)rh$ , how do you solve for r?

Answer 1

In this question, we have to solve the given equation.
First, we need to analyze the equation. Then we can see it is a quadratic equation, so we will evaluate the discriminant. After that, solving this equation by Quadratic formula, we will get the required solution.

Formula used: Quadratic formula:
The roots of the quadratic equation $a{x^2} + bx + c = 0$ is given by –
$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$

Complete step by step solution:
It is given that, $A = 2\left( \pi \right){r^2} + 2\left( \pi \right)rh$ ……. i)
We need to solve the equation for r.
After noticing the equation, we can easily say that it is a quadratic equation in terms of r. We can write the equation as,
$2\left( \pi \right){r^2} + 2\left( \pi \right)rh - A = 0$ ……… ii)
On comparing the given equation with the standard quadratic equation $a{r^2} + br + c = 0$ , we get the following values,
$a = 2\pi$ , $b = 2\pi h$ and $c = - A$
Therefore, solving the equation $2\left( \pi \right){r^2} + 2\left( \pi \right)rh - A = 0$ by Quadratic formula, we get,
$r = \dfrac{{ - 2\pi h \pm \sqrt {{{\left( {2\pi h} \right)}^2} - \left( {4 \times 2\pi \times - A} \right)} }}{{2 \times 2\pi }}$
Or, $r = \dfrac{{ - 2\pi h \pm \sqrt {4{\pi ^2}{h^2} + 8A\pi } }}{{4\pi }}$
Solving, we get,
$r = - \dfrac{{2\pi h}}{{4\pi }} \pm \dfrac{{\sqrt {4{\pi ^2}{h^2} + 8A\pi } }}{{4\pi }}$
In the second fraction, we can rewrite the denominator from $4\pi$ to $\sqrt {16{\pi ^2}}$. Thus we can write,
$r = - \dfrac{h}{2} \pm \sqrt {\dfrac{{4{\pi ^2}{h^2} + 8A\pi }}{{16{\pi ^2}}}}$
Or, $r = - \dfrac{h}{2} \pm \sqrt {\dfrac{{4{\pi ^2}{h^2}}}{{16{\pi ^2}}} + \dfrac{{8A\pi }}{{16{\pi ^2}}}}$
Or, $r = - \dfrac{h}{2} \pm \sqrt {\dfrac{{{h^2}}}{4} + \dfrac{A}{{2\pi }}}$

Hence, solving the equation, $A = 2\left( \pi \right){r^2} + 2\left( \pi \right)rh$ for r , we get, $r = - \dfrac{h}{2} \pm \sqrt {\dfrac{{{h^2}}}{4} + \dfrac{A}{{2\pi }}}$ .

Note: Quadratic equation:
In algebra, a quadratic equation is any equation that can be rearranged in standard form as
$a{x^2} + bx + c = 0$ , where $x$ represents an unknown and $a$ , $b$ and $c$ represent known numbers where $a \ne 0$ .
If $a = 0$ , then it will become a linear equation not quadratic as there is no $a{x^2}$ term.
The discriminant is the part of the quadratic formula under the square root.
$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
The discriminant of the quadratic equation $a{x^2} + bx + c = 0$ is
$\Delta = {b^2} - 4ac$
A positive discriminant indicates that the quadratic equation has two distinct and real roots.
A discriminant of zero indicates that the quadratic equation has equal real roots.
A negative discriminant indicates that the quadratic equation has distinct and complex roots.