Video Transcript
Find 𝐴 and 𝐵 such that two 𝑥
over 𝑥 minus three squared is equal to 𝐴 over 𝑥 minus three plus 𝐵 over 𝑥 minus
three squared.
In order to answer this question,
we need to recall how we split a term in the form 𝑝𝑥 plus 𝑞 over 𝑥 minus 𝑎
squared into partial fractions. This term can be written in the
form 𝐴 over 𝑥 minus 𝑎 plus 𝐵 over 𝑥 minus 𝑎 squared as shown in the
question. The value of 𝑝 in our question is
two, 𝑞 is equal to zero, and lowercase 𝑎 is equal to three.
We begin by multiplying all three
of our terms by 𝑥 minus three squared. At this point, we can cancel common
terms on the numerator and denominator. The left-hand side cancels to
become two 𝑥. On the first term on the right-hand
side, we can cancel 𝑥 minus three, leaving us 𝐴 multiplied by 𝑥 minus three. The final term is 𝐵 as the 𝑥
minus three all squared cancel. Our equation is, therefore, two 𝑥
equals 𝐴 multiplied by 𝑥 minus three plus 𝐵.
There are two traditional methods
for calculating 𝐴 and 𝐵 from this point: firstly, by comparing coefficients and
secondly, by substitution. Let’s firstly solve the equation by
comparing coefficients. Distributing the parentheses or
expanding the brackets on the right-hand side gives us 𝐴𝑥 minus three 𝐴. This gives us two 𝑥 is equal to
𝐴𝑥 minus three 𝐴 plus 𝐵. We can now consider the 𝑥 terms on
both sides of the equation and compare the coefficients. On the left-hand side, we have two
𝑥 and on the right-hand side, 𝐴𝑥. Therefore, two is equal to 𝐴. We noticed that there is no
constant on the left-hand side. But on the right-hand side, we have
negative three 𝐴 plus 𝐵. This means that zero is equal to
negative three 𝐴 plus 𝐵.
Substituting in our value for 𝐴
gives us zero is equal to negative three multiplied by two plus 𝐵. Negative three multiplied by two is
equal to negative six. Adding six to both sides of this
equation gives us six is equal to 𝐵. Using partial fractions and then
comparing coefficients, we have found that our values for 𝐴 and 𝐵 are two and six,
respectively.
We’ll now look at the second method
involving substitution. We return to the equation two 𝑥 is
equal to 𝐴 multiplied by 𝑥 minus three plus 𝐵. We can substitute any two values
into this equation to calculate 𝐴 and 𝐵. However, if we substitute 𝑥 equals
three, this will eliminate the parentheses. On the left-hand side, we have two
multiplied by three and on the right-hand side, 𝐴 multiplied by three minus three
plus 𝐵. As three minus three is equal to
zero, the 𝐴 term cancels. Two times three is equal to
six. Therefore, once again, we have the
solution 𝐵 is equal to six.
As 𝑥 minus three was the only
parentheses, we can now choose any integer to substitute into the equation. It is usually sensible to choose
zero or one. If 𝑥 is equal to zero, the
left-hand side will equal zero. On the right-hand side, we have 𝐴
multiplied by zero minus three plus 𝐵. This simplifies to zero is equal to
negative three 𝐴 plus 𝐵. As 𝐵 is equal to six, we can
substitute in this value. Adding three 𝐴 to both sides gives
us three 𝐴 is equal to six. Finally, dividing both sides by
three gives us 𝐴 is equal to two. Once again, this gives us the same
answer as our first method.
The answers to this question are 𝐴
is equal to two and 𝐵 is equal to six.
