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saurya_s
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hardworker_indian
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Bhai
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saurya_s
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Sorry guys, I have justt edited the Q
S
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Alex_NL
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I would say C.

st.1
x can be only 4 (I tried some values, so I hope this is correct).
Alone not sufficient

st.2
y is 4
Alone not sufficient

Together we can find the value of xy.

Please correct me if I am wrong.


Regards,

Alex
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saurya_s
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Hi ALex,

I was wondering what values did you try? Is there any way to limit the trial numbers?
S
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amernassar
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For the first one

x^3-3x^2-2x-8 = 0

(x-4)(x^2 +x+2) = 0

x = 4 or x^2+x+2 = 0 this quadratic equation has no solution since the discriminant is negative

so Our only solution is x = 4


That is why C is the right answer
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srijay007
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I got C as well

same mtd as amernassar



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