Let x , y and z be complex numbers such that
x + y + z = 2
x 2 + y 2 + z 2 = 3
x y z = 4 .
Evaluate
∣ x y + z − 1 1 + y z + x − 1 1 + z x + y − 1 1 ∣
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Are you kidding me! I just spent half an hour finding each variable! It turns out there are six of them! I checked each one by plugging it into the expression, and do you know what I got! I got 0.09859709458272745784920173027593720179274936759355337365. I even checked that answer twice! I have the feeling that there are more than just this solution, after all, they are complex variables.
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If something takes you a half hour to solve for the variables, chances are that there's a quick way to do it.
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Yeah... it's just that it takes forever because all of them are weird decimals and I was going for an EXACT answer... :(
By Vieta's, the set { x , y , z } are roots to the equation A 3 − 2 A 2 + 2 A − 4 = 0 .
The solution is unique, and you will get 6 solutions up to permutation (assuming 3 distinct roots of the cubic).
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Oh, that's why.
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@Finn Hulse – No, that's not a reason to find something different.
Calvin Lin has already written the equation, so the approximate solutions are x ≈ 2 . 4 5 8 4 , y ≈ − 0 . 2 2 9 2 + 1 . 2 5 4 8 i and z ≈ − 0 . 2 2 9 2 − 1 . 2 5 4 8 i . So:
x y + z − 1 1 ≈ − 0 . 2 7 3 2 − 0 . 2 7 8 8 i
y z + x − 1 1 ≈ 0 . 3 2 4 1
z x + y − 1 1 ≈ − 0 . 2 7 3 2 + 0 . 2 7 8 8 i
x y + z − 1 1 + y z + x − 1 1 + z x + y − 1 1 ≈ − 0 . 2 2 2 2
∣ ∣ ∣ x y + z − 1 1 + y z + x − 1 1 + z x + y − 1 1 ∣ ∣ ∣ ≈ 0 . 2 2 2 2
Because the expression is symmetric, every other combination of the roots gives exactly the same fractions and the same result :)
you are wrong
you just trolled yourself :P
Nice explanation, I found each variable but still got it.
I expanded the expression and started substituting for x + y + z , x y z , etc... Took me 40 minutes, but got the same result!
Brillianťé.AbsolutelyBrillianťé
Sharky Kesa, you are a genius................................
Nice One
Thanks. Very good solution.
I got the answer 0.9 .
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Let S be the desired value. Note that
x y + z − 1 = x y + 1 − x − y = ( x − 1 ) ( y − 1 )
from substitution. Likewise,
y z + z − 1 = ( y − 1 ) ( z − 1 )
and
z x + y − 1 = ( z − 1 ) ( x − 1 ) .
Hence,
S = ∣ ( x − 1 ) ( y − 1 ) 1 + ( y − 1 ) ( z − 1 ) 1 + ( z − 1 ) ( x − 1 ) 1 ∣
S = ∣ ( x − 1 ) ( y − 1 ) ( z − 1 ) x + y + z − 3 = ( x − 1 ) ( y − 1 ) ( z − 1 ) − 1 ∣
S = ∣ x y z − ( x y + y z + z x ) + x + y + z − 1 − 1 ∣
S = ∣ 5 − ( x y + y z + z x ) − 1 ∣
But, 2 ( x y + y z + z x ) = ( x + y + z ) 2 − ( x 2 + y 2 + z 2 ) = 1 . This means x y + y z + z x = 2 1 . I we input this value into the equation, we get an answer of
S = ∣ 5 − 2 1 − 1 ∣ = ∣ 9 − 2 ∣ = 0 . 2 2 2 . . .