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- Feb 14, 2012

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- Thread starter anemone
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- #1

- Feb 14, 2012

- 3,920

- Mar 31, 2013

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you mean ab

Let other 2 roots be c and d

From viete’s relation

We have

a+b+ c+d = - 1 ..1

ab+ac+ad+bc+ bd + cd = 0 ..2

abc+abd+acd+bcd = 0 ..3

abcd = - 1 ..4

now letting s = a+ b, t = c+d, p = ab q = cd we get

s+ t = - 1 or t = -1 –s (1)

p + q + st = 0 …(2)

pt + sq = 0 … (3)

pq = -1 or q = - 1/p

now from (2) p – 1/p –s – s^2 = 0 …(5)

and from (3) –p – ps – s/p =0 => s = - p^2/(p^2+1) …(6)

putting in (5) the value of s we get

p – 1/p +p^2/(p^2+ 1) – p^4/(p^2+1)^2 = 0

or multiplying by (p^2+1) we get

p^6 + p^4 +p^3 – p^2 – 1= 0

so p or ab is a root of it