Completing the square in a denominator
Matthew Barrera 
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,
$\endgroup$ 12 Answers
$\begingroup$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
$$\frac{-5}{x^2-4x+20}=\frac{\frac{-5}{16}}{\frac{(x-2)^2}{16}+\frac{16}{16}}$$ $$=\frac{\frac{-5}{16}}{\frac{(x-2)^2}{4^2}+1}$$ $L=-\frac{5}{16}$, $r=2$,$s=4$
$\endgroup$ $\begingroup$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.
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