If frac{x^2-3x+2}{(x-4)(x-3)^2}=frac{A}{x-4}+ | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Given the partial fraction decomposition: $\frac{x^2-3x+2}{(x-4)(x-3)^2} = \frac{A}{x-4} + \frac{B}{x-3} + \frac{C}{(x-3)^2}$.Multiply both sides

Vedclass · Accepted Answer

$-1$

Vedclass · Answer

(C) Given the partial fraction decomposition: $\frac{x^2-3x+2}{(x-4)(x-3)^2} = \frac{A}{x-4} + \frac{B}{x-3} + \frac{C}{(x-3)^2}$.Multiply both sides by $(x-4)(x-3)^2$ to get: $x^2-3x+2 = A(x-3)^2 + B(x-3)(x-4) + C(x-4)$.To find $A$, set $x=4$: $4^2 - 3(4) + 2 = A(4-3)^2 \Rightarrow 16 - 12 + 2 = A(1)^2 \Rightarrow A = 6$.To find $C$, set $x=3$: $3^2 - 3(3) + 2 = C(3-4) \Rightarrow 9 - 9 + 2 = C(-1) \Rightarrow 2 = -C \Rightarrow C = -2$.To find $B$, compare the coefficients of $x^2$ on both sides: $1 = A + B$. Since $A=6$, we have $1 = 6 + B \Rightarrow B = -5$.Finally, calculate $A+B+C = 6 + (-5) + (-2) = 6 - 7 = -1$.

If $\frac{x^2-3x+2}{(x-4)(x-3)^2}=\frac{A}{x-4}+\frac{B}{x-3}+\frac{C}{(x-3)^2}$ then $A+B+C=$

Similar Questions

Which of the following is a partial fraction of $\frac{-x^{2} + 6x + 13}{(3x + 5)(x^{2} + 4x + 4)} =$

If $\frac{2 x+7}{\left(x^2+4\right)\left(x^2+9\right)\left(x^2+16\right)}=\frac{A x+1}{x^2+4}+\frac{B x+m}{x^2+9}+\frac{C x+n}{x^2+16}$,then $\frac{1}{A}+\frac{1}{B}+\frac{1}{C}=$

If $\frac{3x + a}{x^2 - 3x + 2} = \frac{A}{x - 2} - \frac{10}{x - 1}$,then:

If $\frac{x^3}{(2 x-1)(x+2)(x-3)} = A + \frac{B}{2 x-1} + \frac{C}{x+2} + \frac{D}{x-3}$,then $A$ is equal to

If $\frac{2x + 3}{(x + 1)(x - 3)} = \frac{a}{x + 1} + \frac{b}{x - 3}$,then find the value of $a + b$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module