Question

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

Answer 1

Hint: In this question we have to find the value of the expression when \[x = 6\], so we input the given value inside the given polynomial, thus substitute the value 6 in place of \[x\] in the polynomial given in the question and further simplification using the operations addition, subtraction and multiplication we will get the required value.

Complete step-by-step answer:
Given equation is \[A = 2\left( \pi \right){r^2} + 2\left( \pi \right)rh\] ,
Now transform the equation by taking all terms to one side and make the equation equal to 0, we get,
\[2\left( \pi \right){r^2} + 2\left( \pi \right)rh - A = 0\],
Now it is in form a quadratic equation in terms of \[r\], we have to solve the equation using quadratic formula which is given by\[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\],
Here \[x = r\],\[a = 2\pi \],\[b = 2\pi h\] and \[c = - A\],
Now substituting the values in the formula we get,
\[ \Rightarrow r = \dfrac{{ - 2\pi h \pm \sqrt {{{\left( {2\pi h} \right)}^2} - 4\left( {2\pi } \right)\left( { - A} \right)} }}{{2\left( {2\pi } \right)}}\],
Now simplifying we get,
\[ \Rightarrow r = \dfrac{{ - 2\pi h \pm \sqrt {4{\pi ^2}{h^2} + 8\left( \pi \right)\left( A \right)} }}{{4\pi }}\],
Again simplifying by dividing the expression into two parts we get,
\[ \Rightarrow r = \dfrac{{ - 2\pi h}}{{4\pi }} \pm \dfrac{{\sqrt {4{\pi ^2}{h^2} + 8\pi A} }}{{4\pi }}\],
Now simplifying we get,
\[ \Rightarrow r = \dfrac{{ - h}}{2} \pm \dfrac{{\sqrt {4{\pi ^2}{h^2} + 8\pi A} }}{{4\pi }}\],
Now taking the denominator into the square root by squaring the denominator we get,
\[ \Rightarrow r = \dfrac{{ - h}}{2} \pm \sqrt {\dfrac{{4{\pi ^2}{h^2} + 8\pi A}}{{16{\pi ^2}}}} \],
Now again simplifying the part in the square root we get,
\[ \Rightarrow r = \dfrac{{ - h}}{2} \pm \sqrt {\dfrac{{4{\pi ^2}{h^2}}}{{16{\pi ^2}}} + \dfrac{{8\pi A}}{{16{\pi ^2}}}} \],
Further simplification we get,
\[ \Rightarrow r = \dfrac{{ - h}}{2} \pm \sqrt {\dfrac{{{h^2}}}{{4{\pi ^2}}} + \dfrac{A}{{2\pi }}} \],
So the value of \[r\] is equal to \[\dfrac{{ - h}}{2} \pm \sqrt {\dfrac{{{h^2}}}{{4{\pi ^2}}} + \dfrac{A}{{2\pi }}} \].

Final Answer:
\[\therefore \]The value of the value of the equation in terms of \[r\] is equal to \[\dfrac{{ - h}}{2} \pm \sqrt {\dfrac{{{h^2}}}{{4{\pi ^2}}} + \dfrac{A}{{2\pi }}} \].

Note:
We have to remember that in linear equations and from some basic properties, that when we are asked for a variable we have to treat all other terms as constants. If we treat \[A,h,\pi \] as a variable then the given equation will become 5 variables which will have infinite solutions.