Question

25 point for this question:

For which values of a and b the following is true for any real number x?
$(x-4)/x^2+7x-18=a/x+9+b/x-2$

Answer 1

Answer:

a=$\frac{13}{11}$   b =$\frac{-2}{11}$

Step-by-step explanation:

• By equating coefficients on both sides of respective terms.

Given,

$\frac{x-4}{x^2+7x-18}$ = $\frac{a}{x+9}$ + $\frac{b}{x-2}$

Now take  lcm on right hand side

$\frac{x-4}{x^2+7x-18}$ =$\frac{a(x-2)+b(x+9)}{x^2+7x-18}$

By equating coefficients of respective terms

x-4 = a(x-2)+b(x+9)

a+b=1  -----1 ; 9b-2a=-4  -----2

Substitute b=1-a in 2

9(1-a)-2a=-4

11a=9+4=13

a=$\frac{13}{11}$

As b=1-a=1- $\frac{13}{11}$=$\frac{-2}{11}$