Why doesn’t the equation "break" when we put a value $x$ while solving partial fractions?

Question

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?

Answer 1

If you write $$ \frac{9}{9-x^2}=\frac{a}{3-x}+\frac{b}{3+x} $$ you get $$ \frac{9}{9-x^2}=\frac{a(3+x)+b(3-x)}{9-x^2} $$ This means that, for all $x\ne\pm3$, we must have $$ a(3+x)+b(3-x)=9 $$ Since the left-hand side is a continuous function for all $x$, the equality also holds for $x=\pm3$.

For $x=3$, $6a=9$; for $x=-3$, $6b=9$.

Answer 2

You are right, that equation doesn't make sense at exactly $x=3$. But let's see what will happen if $x=2.999999$:

$$ \frac{9}{(3-2.999999)(3+2.999999)} = \frac{A}{3-2.999999}+\frac{B}{3+2.999999}\\ 10^6\frac{9}{5.999999}=10^6A+\frac B{5.999999}\\ A = \frac{9}{5.999999} - 0.000001\frac B{5.999999} $$

As $x$ will approach closely and closely to 3, the contribution of $B$ will be smaller and smaller, so in the limit $x\to 3$, $A=9/6=3/2$. Thus the substitution of $x=3$ should be understood in terms of limits and not direct substitution.

Answer 3

One way of looking at this is that we are seeking to equate the numerators of the fractions algebraically (we forget the denominator for the moment). If we can find an identity in the numerators the decomposition will be valid wherever the denominator is non-zero.

To establish such an identity, which is simply an identity between polynomials, and is indicated to exist because there are as many equations between coefficients as there are unknowns, we can use any method we choose. Picking convenient values works as well as any other.

When the denominator is zero, this also applies to one of the component parts of the decomposition, so they represent the same function. But the intermediate step of equating numerators does not involve dividing by zero.

Answer 4

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 $x.$

Answer 5

$$\frac{9}{9-x^2} = \frac{a}{3+x} + \frac{b}{3-x} \tag{A}$$

LHS and RHS are continuous except at the two points $x=3$ and $x=-3$.

Multiplying through by $9-x^2$ gives

$$9 = a(3-x) + b(3+x) \tag{B}$$

where both LHS and RHS are continuous for all real numbers and can be solved for the parameters $a$ and $b$.

Once you have found the values of $a$ and $b$, you can fill in the values for $a$ and $b$ and then divide both sides of equation (B) by $9-x^2$. You will get equation (A) with the values of $a$ and $b$ filled in and equation (A) will be true for all $x$ except at $x=3$ and $x=-3$.