rohan_vus wrote:One thing for sure a cant be +ve or 0..coz question stem says " sum of the roots equals the sum of the squares of their reciprocals"..
Sum of roots = -a ...so a must be -ve...as sum of roots cant be -ve ( given the fact that its sum of squares of reciprocals )
So left with A and E.. putting a = -1 in the original equation doesnt give real roots ,so A is also ruled out
x^2 - x + 1 ===> roots are imaginary...|a| should be >=4 for some real roots...
If question would have been x^2 + ax - 1 then a = -1 would have been possible answer ..but thats not the case here
So E remains
Let's try it like this, rohan
Let's say that α and β be the roots of the quadratic x^2 + a x + 1 = 0, so that
α + β = -a and α β = 1, hence sum of the roots (i.e. -a) = sum of the squares of the reciprocals of roots
or -a = 1/ α^2 + 1/β^2 = (α^2 + β^2)/ α^2 β^2 = {(α + β)^2 - 2 α β}/ α^2 β^2, put values to get
-a = {(-a)^2 - 2 × 1}/(1)^2
» -a = a^2 - 2
» a^2 + a - 2 = 0
» (a - 1) (a + 2) = 0
» a is either 1 or -2.
Is [spoiler](C)[/spoiler] ok here, rohan?
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Sanjeev K Saxena
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The Princeton Review - Manya Abroad
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