2 > If a , b are the roots of x2 + px + q = 0 , and they are also the roots of the eqn.
x2n + pn xn + qn = 0 , then prove that a / b and b / a are the roots of
the eqn. xn + 1 + ( x + 1 )n whenever n is an even integer , and vice versa .
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1 Answers
23a+b = - p (1)
ab = q (2)
a,b are roots of x2n + pn xn + qn = 0
thus a2n + pn an + qn = 0
i.e
(an)2 + pn an + qn = 0
thus we can say that an is a root of the quadratic eqn
x2 + pnx+qn=0
similarly bn is also a root of this eqn
thus an+bn= -pn
anbn = qn
now consider xn + 1 + ( x + 1 )n
put x = b/a and take LCM
thus
\frac{b^{n}}{a^{n}}+1+(\frac{b+1}{a})^{n}
=\frac{b^{n}+ a^{n} +(b+a)^{n}}{a^{n}}
= \frac{-p^{n} + p^{n}}{a^{n}}
(since an+bn = -pn
and a+b = -p
and thus (a+b)n = pn since n = even integer )
=0
similarly we can get the eqn = 0 for x = a/b
