Q.145. To find a complex root of a polynomial.

Question 145.
Find exact values of the real numbers a and b, if a + b i is a root of x^4 - 4x^3 + 48x^2 - 64x + 256 = 0.
There is more than one answer, but any one will suffice.

Answer 145.
x^4 - 4x^3 + 48x^2 - 64x + 256 = 0
=> (x^4 + 32x^2 + 256) - 4x(x^2 + 16) + 4x^2 + 12x^2 = 0
=> (x^2 + 16)^2 - 4x(x^2 + 16) + (2x)^2 - (2√3ix)^2 = 0
=> (x^2 - 2x + 16)^2 - (2√3ix)^2 = 0
=> [x^2 - 2(√3i + 1)x + 16] [x^2 + 2(√3i - 1)x + 16] = 0
=> EITHER x
= (1/2)[2(√3i +1) ± √[4(√3i +1)^2 - 64]
= (√3i + 1) ± √(2√3i - 18)
OR x
= (1/2)[- 2(√3i -1) ± √[4(√3i -1)^2 - 64]
= -(√3i - 1) ± √(-2√3i - 18)

Let one of the four roots to be simplified be
(√3i + 1) + √(2√3i - 18) = a + bi
Let m+ni = √(2√3i - 18)
=> m^2 - n^2 + 2mni = -18 + 2√3i
=> n^2 - m^2 = 18 and mn = √3
=> 3/m^2 - m^2 = 18
=> m^4 + 18m^2 - 3 = 0
=> m^2 = (1/2)[-18 ± √336]
=> m^2 = - 9 + √84
=> m = √(-9 + √84) and n = √(9 + √84)
=> a + bi
= (√3i + 1) + √(-9 + √84) + i√(9 + √84)
= 1 + √(-9 + √84) + i [√3 + √(9 + √84)]
=> a = 1 + √(-9 + √84) and b = √3 + √(9 + √84).

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