Question:
If
\[
\frac{13x+43}{2x^2+17x+30}
=
\frac{A}{2x+5}
+
\frac{B}{x+6},
\]
then \(A^2+B^2=\)
If
\[
\frac{13x+43}{2x^2+17x+30}
=
\frac{A}{2x+5}
+
\frac{B}{x+6},
\]
then \(A^2+B^2=\)
Show Hint
In partial fractions, after taking LCM, compare the coefficients of like powers of \(x\) to find unknown constants.
Updated On: Jun 22, 2026
- \(\dfrac{22}{3}\)
- \(52\)
- \(34\)
- \(\dfrac{18}{5}\)
Show Solution
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The Correct Option is C
Solution and Explanation
Step 1: Factorize the denominator.
We have \[ 2x^2+17x+30 \] Factorizing, \[ 2x^2+17x+30=(2x+5)(x+6) \] So, \[ \frac{13x+43}{2x^2+17x+30} = \frac{13x+43}{(2x+5)(x+6)} \]
Step 2: Write the partial fraction form.
Given, \[ \frac{13x+43}{(2x+5)(x+6)} = \frac{A}{2x+5} + \frac{B}{x+6} \] Taking LCM on the right side, \[ \frac{A}{2x+5} + \frac{B}{x+6} = \frac{A(x+6)+B(2x+5)}{(2x+5)(x+6)} \]
Step 3: Compare numerators.
Thus, \[ 13x+43=A(x+6)+B(2x+5) \] Expanding, \[ 13x+43=Ax+6A+2Bx+5B \] \[ 13x+43=(A+2B)x+(6A+5B) \] Comparing coefficients of \(x\) and constant terms, we get \[ A+2B=13 \] and \[ 6A+5B=43 \]
Step 4: Solve for \(A\) and \(B\).
From \[ A+2B=13, \] we get \[ A=13-2B \] Substitute this in \[ 6A+5B=43 \] \[ 6(13-2B)+5B=43 \] \[ 78-12B+5B=43 \] \[ 78-7B=43 \] \[ -7B=-35 \] \[ B=5 \] Now, \[ A=13-2(5) \] \[ A=3 \]
Step 5: Find \(A^2+B^2\).
\[ A^2+B^2=3^2+5^2 \] \[ =9+25 \] \[ =34 \]
Step 6: Final conclusion.
Hence, \[ \boxed{34} \] which corresponds to option (3).
We have \[ 2x^2+17x+30 \] Factorizing, \[ 2x^2+17x+30=(2x+5)(x+6) \] So, \[ \frac{13x+43}{2x^2+17x+30} = \frac{13x+43}{(2x+5)(x+6)} \]
Step 2: Write the partial fraction form.
Given, \[ \frac{13x+43}{(2x+5)(x+6)} = \frac{A}{2x+5} + \frac{B}{x+6} \] Taking LCM on the right side, \[ \frac{A}{2x+5} + \frac{B}{x+6} = \frac{A(x+6)+B(2x+5)}{(2x+5)(x+6)} \]
Step 3: Compare numerators.
Thus, \[ 13x+43=A(x+6)+B(2x+5) \] Expanding, \[ 13x+43=Ax+6A+2Bx+5B \] \[ 13x+43=(A+2B)x+(6A+5B) \] Comparing coefficients of \(x\) and constant terms, we get \[ A+2B=13 \] and \[ 6A+5B=43 \]
Step 4: Solve for \(A\) and \(B\).
From \[ A+2B=13, \] we get \[ A=13-2B \] Substitute this in \[ 6A+5B=43 \] \[ 6(13-2B)+5B=43 \] \[ 78-12B+5B=43 \] \[ 78-7B=43 \] \[ -7B=-35 \] \[ B=5 \] Now, \[ A=13-2(5) \] \[ A=3 \]
Step 5: Find \(A^2+B^2\).
\[ A^2+B^2=3^2+5^2 \] \[ =9+25 \] \[ =34 \]
Step 6: Final conclusion.
Hence, \[ \boxed{34} \] which corresponds to option (3).
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