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"..till you consider cases for imaginary roots.. But in GMAT you are not tested for imaginary roots as such..so on these lines rest follows

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

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?

a could be -2 but not 1... Even if you get 2 values of a as 1 and -2 still , you are not sure which value shall give you real roots ( till you care about imaginary roots)

with a = 1 , your eqn is x^2 + x + 1 = 0 gives imaginary cube roots of unity which are w and w^2 -( w = omega)
with a = -2 you get x^2 - 2x + 1 = 0 ==> x = 1..Here its perfect ...

But a = -2 is not listed in the choices given

rohan_vus wrote:a could be -2 but not 1... Even if you get 2 values of a as 1 and -2 still , you are not sure which value shall give you real roots ( till you care about imaginary roots)

with a = 1 , your eqn is x^2 + x + 1 = 0 gives imaginary cube roots of unity which are w and w^2 -( w = omega)
with a = -2 you get x^2 - 2x + 1 = 0 ==> x = 1..Here its perfect ...

But a = -2 is not listed in the choices given

On GMAT, we can still expect a quadratic that leads to imaginary roots, but the question won't be about finding the roots of those particular quadratics. Simply because GMAT never involves imaginary numbers in its quantitative discussions. Here we are discussing about the sum and product of the roots (whatever) of the possible quadratic x^2 + x + 1 = 0 when a = 1, which are coming out to be real numbers like -1 and 1. I think that's kewl, rohan!!

Ok !!!

Btw , there are many ways as such to solve such cases w/o even getting to writing equations .Intuitively one can eliminate choices based on what i was discussing ,where you dont even need to do any pen work.Mine was one way to do illustrate that.But unfortunately this method goes with assumption that roots cant be imaginary .

If answer choices would have contained -2 for e.g you would have had your pick straight away by the above method .But here that was not the case .

Other way , Conventional one, one can indeed arrive at what you did and certainly safe bet at times