Mathematics Question
x^3 + 4x +3=0 has roots (alpha),(beta),(gamma)
using the substitution x=(square root)u
find the value of (alpha)^4 + (beta)^4 + (gamma)^4 +(alpha)(beta)(gamma)
The answer is 29. Let x1, x2, x3 be the roots. We have a3x^3+a2x^2+a1x+a0=0; In our case a3=1, a2=0, a1=4, a0=3; Viet's theorem gives: x1+x2+x4=-a2=0; x1x2+x2x3+x1x3=a1=4; x1x2x3=-a0=-3; So, 0=(x1+x2+x3)^2=x1^2+x2^2+x3^2+2(x1x2+x2 Which gives: x1^2+x2^2+x3^2 = -8; Now if we raise that to the power of 2: (-8)^2 = 64 =(x1^2+x2^2+x3^2)^2 =x1^4+x2^4+x3^4 + 2((x1x2)^2+(x2x3)^2+(x1x3)^2); The value in brackets is: (x1x2)^2+(x2x3)^2+(x1x3)^2=(x1x2+x2x3+x (x1x2 + x2x3 + x1x3)^2, since x1+x2+x3=0; So, finally: 64=x1^4 + x2^4 + x3^4 +2(x1x2 + x2x3 + x1x3)^2; 64 = x1^4 + x2^4 + x3^4 +2*4^2; x1^4 + x2^4 + x3^4 = 32 And the answer: x1^4 + x2^4 + x3^4 + x1x2x3 = 32 + (-3) = 29. 1
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well you pretty much used the right method using Viet's theorem, but the way you've chosen to do it you end up getting unwanted terms such as 2(x1)(x2)(x3^2) when you raise x1x2+x2x3+x1x3 to the 2nd power which i'm not sure you accounted for and would need a lot more work to get rid of. 1
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