Question

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

Answer 1

Hint: 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.