+0

# Find constants A and B

0
62
4

Find constants A and B such that

(x+7)/(x²-x-2)+a/(x-2)+b/(x+1)

for all x such that x is not equal to -1 and x is not equal to 2. Give your answer as the ordered pair (A,B).

Aug 14, 2020

#1
+1

Notice that x-2 and x+1 are factors of x^2-x-2.

$$\frac{x+7}{x^2-x-2}=\frac{a}{x-2}+\frac{b}{x+1}\\ \frac{x+7}{x^2-x-2}=\frac{a(x+1)}{x^2-x-2}+\frac{b(x-2)}{x^2-x-2}\\ x-7=a(x+1)+b(x-2)\\ x-7=ax+a+bx-2b\\$$

ax+bx must add up to the x term on the left side, which is x. a-2b must subtract to the constant on the left side, -7. We then have a system of equations.

a+b=1

a-2b=-7

We find that a= -5/3 and b=8/3.

Aug 14, 2020
#2
0

Guest Aug 14, 2020
#3
+110762
+1

$$\frac{(x+7)}{(x²-x-2)}+\frac{a}{x-2}+\frac{b}{(x+1)}$$

Such that what?  Is there meant to be an equal sign somewhere?

Aug 15, 2020
#4
+30931
+2

I assume this is a partial fraction question as interpreted by Guest#1.  Guest#1 has the right approach, but unfortunately mistakenly set a - 2b = -7, where it should be a - 2b = 7.  Given this correction you shoud be able to determine a and b.

Aug 15, 2020