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Consider the polynomial f(x)=17x4+21x3+60x2+Ax+B. Suppose that for every root λ of f(x)=0, 1λ is also a root of f(x)=0. What is...Asked by Anonymous1
Consider the polynomial f(x)=17x4+21x3+60x2+Ax+B. Suppose that for every root λ of f(x)=0, 1/λ is also a root of f(x)=0. What is the value of A+B?
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Answered by
Steve
If two of the roots are m and n,
f(x) = (x-m)(x-n)(mx-1)(nx-1)
B = mn
17 = mn
so, B=17
mn*1/m + mn*1/n + m*1/m*1/n + n*1/m*1/n = -A/17
n + m + 1/m + 1/n = -A/17
But, we also know that the sum of the roots is -21/17, so
A = 21
A+B = 38
As it happens, all he roots are complex, but they are reciprocals in pairs!
f(x) = (x-m)(x-n)(mx-1)(nx-1)
B = mn
17 = mn
so, B=17
mn*1/m + mn*1/n + m*1/m*1/n + n*1/m*1/n = -A/17
n + m + 1/m + 1/n = -A/17
But, we also know that the sum of the roots is -21/17, so
A = 21
A+B = 38
As it happens, all he roots are complex, but they are reciprocals in pairs!
Answered by
Steve
Hmmm. Getting B is ill written.
B/17 = product of all roots
m*n*1/m*1/n = 1
B/17 = 1
B = 17
B/17 = product of all roots
m*n*1/m*1/n = 1
B/17 = 1
B = 17
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