Let’s suppose we want to decompose $\frac {9}{9-x^2}$ by partial fractions. I was taught that we proceed by writing
$$ \frac{9}{9-x^2} = \frac{a}{3+x} + \frac{b}{3-x}, $$
where $a$ and $b$ are unknown constants. Clearing the fractions, we get
$$(3-x)a + (3+x)b = 9.$$
To find $a$, we assume that $x=-3$ and solve the resulting equation; and similarly, to find $b$, we assume that $x=3$ and solve. But when we assume it we are diving by $0$ in the original fraction which isn't possible!
I am confused. How does this procedure give the correct result?
Best Answer
You began by supposing that $$ \frac{9}{9-x^2}=\frac{a}{3-x}+\frac{b}{3+x}. \tag1 $$
Now you need to find values of $a$ and $b$ that make this true for all $x$ such that $x \neq 3$ and $x \neq -3.$
Next you discover that for every $x$ you care about (every possible value of $x$ except $3$ and $-3$), Equation $(1)$ is equivalent to $$ 9= a(3+x) + b(3-x). \tag2 $$
If you can find values of $a$ and $b$ that make Equation $(2)$ true for all $x$ such that $x \neq 3$ and $x \neq -3,$ then you have solved for $a$ and $b$ in Equation $(1)$.
But suppose you can find $a$ and $b$ that Equation $(2)$ true, not just for all $x$ such that $x \neq 3$ and $x \neq -3,$ but for every possible value of $x$ including the possible values $x = 3$ and $x = -3.$ Does the fact that the equation happens to be true for $x=3$ and for $x = -3,$ rather than undefined, give us any a priori reason to doubt that it is true for other values of $x$?
Plugging in $x = 3$ and $x = -3,$ and solving for $a$ and $b$ in each case does not actually (by itself) prove anything about the other values of $x.$ But it does tell us values of $a$ and $b$ that make Equation $(2)$ true for those values of $x$: $a = b = \frac32.$
Now look what happens if you plug these values into the right-hand side of Equation $(2)$: \begin{align} a(3+x) + b(3-x) &= \frac32(3+x) + \frac32(3-x) \\ &= \frac32(3) + \frac32 x + \frac32(3) - \frac32 x \\ &= 9 \end{align} for every possible real number $x,$ including all the possible values that are not $3$ or $-3.$
That's how it works. We really did not care what values of $a$ and $b$ make $9= a(3+x) + b(3-x)$ when $x = 3$ or when $x = -3.$ Setting $x$ to those values is just a technique for finding out what we need $a$ and $b$ to be in order to make $9= a(3+x) + b(3-x)$ for all the other possible values of $x.$ And once we have that, we know that $\frac{9}{9-x^2}=\frac{a}{3-x}+\frac{b}{3+x}$ also will be true for all those values of $x.$