TS EAMCET · Maths · Quadratic Equation
For \(n>2\) and \(n \in \mathbf{N}\), the product of the roots of \((x-n)\left(\left(x^2-2 n x\right)^2+\left(2 n^2-5\right)\left(x^2-2 n x\right)\right.\) \(\left.+\left(n^4-5 n^2+4\right)\right)=0\) is divisible by
- A 625
- B 25
- C 120
- D 80
Answer & Solution
Correct Answer
(C) 120
Step-by-step Solution
Detailed explanation
Given that, \((x-n)\left(\left(x^2-2 n x\right)^2+\left(2 n^2-5\right)\left(x^2-2 n x\right)\right.\) \(\left.+\left(n^4-5 n^2+4\right)\right)=0\) \(\Rightarrow(x-n)\left(x^4-4 n x^3+4 n^2 x^2+2 n^2 x^2-4 n^3 x\right.\) \(\left.-5 x^2+10 n x+n^4-5 n^2+4\right)=0\)…
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