If frac{13x+43}{2x^2+17x+30} = frac{A}{2x+5} | vedclass

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

Question

(A) Given the partial fraction decomposition: $\frac{13x+43}{2x^2+17x+30} = \frac{A}{2x+5} + \frac{B}{x+6}$ Combining the terms on the right side: $\f

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(A) Given the partial fraction decomposition: $\frac{13x+43}{2x^2+17x+30} = \frac{A}{2x+5} + \frac{B}{x+6}$ Combining the terms on the right side: $\frac{13x+43}{2x^2+17x+30} = \frac{A(x+6) + B(2x+5)}{(2x+5)(x+6)}$ Since the denominators are equal,we equate the numerators: $13x + 43 = A(x+6) + B(2x+5)$ Expanding the right side: $13x + 43 = (A + 2B)x + (6A + 5B)$ Comparing the coefficients of $x$ and the constant terms,we get the system of equations: $A + 2B = 13$ ... $(i)$ $6A + 5B = 43$ ... $(ii)$ From $(i)$,$A = 13 - 2B$. Substituting this into $(ii)$: $6(13 - 2B) + 5B = 43$ $78 - 12B + 5B = 43$ $-7B = 43 - 78$ $-7B = -35 \Rightarrow B = 5$ Substituting $B = 5$ back into $(i)$: $A + 2(5) = 13 \Rightarrow A + 10 = 13 \Rightarrow A = 3$ Therefore,$A + B = 3 + 5 = 8$.

If $\frac{13x+43}{2x^2+17x+30} = \frac{A}{2x+5} + \frac{B}{x+6}$,then $A+B = $

Similar Questions

If $\frac{x^4}{(x-1)(x-2)(x-3)}=p(x)+\frac{A}{x-1}+\frac{B}{x-2}+\frac{C}{x-3}$ then $p\left(\frac{3}{2}\right)+C=$

If $\frac{1}{x^4+x^2+1}=\frac{A x+B}{x^2+x+1}+\frac{C x+D}{x^2-x+1}$,then $C+D$ is equal to

If $F_1$ and $F_2$ are irreducible factors of $x^4+x^2+1$ with real coefficients and $\frac{x^3-2x^2+3x-4}{x^4+x^2+1}=\frac{Ax+B}{F_1}+\frac{Cx+D}{F_2}$,then $A+B+C+D=$

If $\frac{x^2}{2 x^4+7 x^2+6}=\frac{A x+B}{x^2+a}+\frac{C x+D}{a x^2+3}$, then find the value of $A+B+C-2 D$. (in $a$)

The partial fraction of $\frac{3x - 1}{(1 - x + x^2)(2 + x)}$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module