TS EAMCET · Maths · Quadratic Equation
Let \(\alpha \neq 1\) be a real root of the equation \(x^3-a x^2+a x-1=0\), where \(a \neq-1\) is a rea number. Then, a root of this equation, among the following, is
- A \(\alpha^2\)
- B \(-\frac{1}{\alpha}\)
- C \(\frac{1}{\alpha}\)
- D \(-\frac{1}{\alpha^2}\)
Answer & Solution
Correct Answer
(C) \(\frac{1}{\alpha}\)
Step-by-step Solution
Detailed explanation
Equation \(x^3-a x^2+a x-1=0\) and \(\alpha \neq 1\), \(a \neq-1\). we put, \(\left(x=\frac{1}{y}\right)\) in given equation Then, \(\left(\frac{1}{y}\right)^3-a\left(\frac{1}{y}\right)^2+a\left(\frac{1}{y}\right)-1=0\) \(\frac{1}{y^3}-\frac{a^2}{y^2}+\frac{a}{y}-1=0\)…
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