### Instruction: When the diameter of a pizza increases by 2 inches, the area increases by $44\%$. What was the area, in square inches, of the original pizza? Express your answer in terms of $\pi$.### Response: The area of a pizza is given by the formula $A=\pi r^2$, where $r$ is the radius of the pizza.
If the diameter increases by 2 inches, then the radius increases by 1 inch.
So the new radius is $r+1$ and the new area is $A'=\pi (r+1)^2$.
The increase in area is $44\%$ of the original area, so we can write the equation $A'=A+0.44A$.
Substituting the expressions for $A$ and $A'$, we get $\pi (r+1)^2=\pi r^2+0.44\pi r^2$.
Simplifying, we get $(r+1)^2=r^2+0.44r^2$, or $r^2+2r+1=1.44r^2$.
Rearranging, we get $0.44r^2-2r-1=0$.
To solve for $r$, we can use the quadratic formula.
We have $a=0.44$, $b=-2$, and $c=-1$.
The quadratic formula gives $r=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$.
Plugging in the values, we get $r=\frac{2\pm\sqrt{4-4(0.44)(-1)}}{2(0.44)}$.
Simplifying, we get $r=\frac{2\pm\sqrt{5.76}}{0.88}$.
Now, we have to choose the positive value of $r$, since a negative radius does not make sense.
So we take $r=\frac{2+\sqrt{5.76}}{0.88}$.
The original area of the pizza is $A=\pi r^2$, so we plug in the value of $r$ and get $A=\pi \left(\frac{2+\sqrt{5.76}}{0.88}\right)^2$.