TS EAMCET · Maths · Quadratic Equation
If \(\frac{27 x^2+32 x+16}{\beta x+2)^2(1-x)}=\frac{A}{3 x+2}+\frac{B}{\beta x+2)^2}\) \(+\frac{C}{1-x}\), then \(A B+B C+C A=\)
- A 6
- B 12
- C 24
- D 48
Answer & Solution
Correct Answer
(B) 12
Step-by-step Solution
Detailed explanation
We have, \[ \begin{aligned} & \frac{27 x^2+32 x+16}{(3 x+2)^2(1-x)}=\frac{A}{3 x+2}+\frac{B}{(3 x+2)^2}+\frac{C}{1-x} \\ & \Rightarrow 27 x^2+32 x+16=A(3 x+2)(1-x)+B \\ & (1-x)+C(3 x+2)^2 \end{aligned} \] Put \(x=1\) both sides, we get…
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