TS EAMCET · Maths · Quadratic Equation
If \(\alpha\) is a root of multiplicity 3 of the equation \(x^5-8 x^4+25 x^3-38 x^2+28 x-8=0\), then \(\alpha^2-5 \alpha+6=\)
- A 0
- B 1
- C 2
- D 3
Answer & Solution
Correct Answer
(A) 0
Step-by-step Solution
Detailed explanation
We have, \(x^5-8 x^4+25 x^3-38 x^2+28 x-8=0\) Let…
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