Completing the square in a denominator

Question

I am given a fraction: $$\frac{J}{(x^2 -4x +20)}$$, where $J=-5$.

I am told that by completing the square in the denominator, and by dividing the top and bottom by a constant I should be able to express the given fraction in the form:

$$\frac{L}{(x-r)^2/s^2+1}$$

, for some constants $L, r, s$, with $s>0$. I am asked to find $L, r$, and $s$.

So far, I have tried to complete the square:

$x^2 -4x +20=0$

$(x-2)^2 = -16$,

however, I am going to take the square root and it results in a complex number which I believe is false, as this fraction will be used later on for integration.

Any help is appreciated, Thank you for your time,

Answer 1

Since by completing the squares we know $$x^2-4x+20=(x-2)^2+16$$ Therefore $$\frac{-5}{x^2-4x+20}=\frac{-5}{(x-2)^2+16}$$ Dividing top and bottom by 16

$$\frac{-5}{x^2-4x+20}=\frac{\frac{-5}{16}}{\frac{(x-2)^2}{16}+\frac{16}{16}}$$ $$=\frac{\frac{-5}{16}}{\frac{(x-2)^2}{4^2}+1}$$ $L=-\frac{5}{16}$, $r=2$,$s=4$

Answer 2

Completing the square is a technique that most students will first encounter when learning how to solve quadratic equations, $ax^2+bx+c=0$. I believe this is the source of the confusion.

In this case you are not being asked to solve a quadratic equation (there is no equals sign), you are being asked to transform a quadratic form $ax^2+bx+c$ into a different format.

The fact that this quadratic form can't be 0 for any real $x$ (as you have found out) is actually a good thing, as it is in the denominator of a fraction, and of course dividing by zero leads to issues / infinities / discontinuities.

If you follow the steps shown in Clark Makmur's answer you obtain the form you have been asked for, without needing to take square roots of any negative numbers.