a b c + a b + b c + c a + a + b + c b c d + b c + c d + d b + b + c + d c d a + c d + d a + a c + c + d + a d a b + d a + a b + b d + d + a + b = = = = 1 7 2 9 There exists a monic quartic polynomial with non-real roots a , b , c , d . Compute its discriminant.
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M y H a r d R o c k M e t h o d :
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http://mathworld.wolfram.com/PolynomialDiscriminant.html
Better:
a = -0.0212830897077841
b = 1.60991176077924
c = -0.217026471766227
d = 2.91486764116886
Worse:
a = -1.48935845514611 ± 0.847593707426474i
b = -2.30495588038962 ± 2.2602498864706i
c = -1.39148676411689 ± 0.67807496594118i
d = -2.95743382058443 ± 3.3903748297059i
Answer: 4 9
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add one to each eqns to find: ( a + 1 ) ( b + 1 ) ( c + 1 ) = 2 ( b + 1 ) ( c + 1 ) ( d + 1 ) = 8 ( c + 1 ) ( d + 1 ) ( a + 1 ) = 3 ( d + 1 ) ( a + 1 ) ( b + 1 ) = 1 0 let A=a+1,B=b+1,C=c+1,D=d+1. then A B C = 2 , B C D = 8 , C D A = 3 , D A B = 1 0 multiply all to get ( A B C D ) 3 = 2 ∗ 8 ∗ 3 ∗ 1 0 = 4 8 0 subtract each eqn simultaneously : ⎩ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎧ ( A − D ) B C = 2 − 8 = − 6 ( A − C ) B D = 1 0 − 8 = 2 ( A − B ) C D = 3 − 8 = − 5 ( B − C ) A D = 1 0 − 3 = 7 ( B − D ) A C = 2 − 3 = − 1 ( D − C ) A B = 1 0 − 2 = 8 multiply all these to get ( A − B ) ( A − C ) ( A − D ) ( B − C ) ( B − D ) ( D − C ) ( A B C D ) 3 = − 3 3 6 0 ( A − B ) ( A − C ) ( A − D ) ( B − C ) ( B − D ) ( D − C ) = − 7 Δ = ( ( a − b ) ( a − c ) ( a − d ) ( b − c ) ( b − d ) ( d − c ) ) 2 = 4 9