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squares of their reciprocals

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by sanju09 » Wed Mar 03, 2010 2:38 am
If the sum of the roots of the equation x^2 + t x + 1 = 0 is equal to the sum of the squares of their reciprocals, then which of the following is a possible value of t?
(A) -1
(B) 0
(C) ½
(D) 1
(E) 2
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by harsh.champ » Wed Mar 03, 2010 4:14 am
sanju09 wrote:If the sum of the roots of the equation x^2 + t x + 1 = 0 is equal to the sum of the squares of their reciprocals, then which of the following is a possible value of t?
(A) -1
(B) 0
(C) ½
(D) 1
(E) 2
Let the roots be a and b.
a+b=-t
ab = 1
Sum of squares of their reciprocals = 1/a^2 + 1/ b^2
=(a^2 + b^2)/(ab)^2
=[(a+b)^2 - 2ab]/(ab)^2
= t^2 - 2
Now, t^2 - 2 = -t
=>t^2 +t - 2 = 0
[spoiler]t= 1 D[/spoiler]
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