How do you find all the real and complex roots of 8x^5-6x^4-83x^2-6x+8=0?
1 Answer
Here's an answer based on the supposition that the equation should have been:
8x^4-6x^3-83x^2-6x+8=0
Explanation:
Let
Due to the symmetry of the coefficients, if
r^4 f(1/r) = r^4 (8r^(-4)-6r^(-3)-83r^(-2)-6r^(-1)+8)
=8-6r-83r^2-6r^3+8r^4 = f(r)
Hence
f(x) = 8(x^2+ax+1)(x^2+bx+1)
=8(x^4+(a+b)x^3+(2+ab)x^2+(a+b)x+1)
=8x^4+8(a+b)x^3+8(2+ab)x^2+8(a+b)x+8
Equating coefficients, we find:
8(a+b) = -6
8(2+ab) = -83
Hence:
a+b = -3/4
ab = -83/8-2 = -99/8
So
t^2+3/4t-99/8 = 0
That is:
8t^2+6t-99 = 0
Use the quadratic formula to find roots:
(-6+-sqrt(6^2-(4*8*-99)))/(2*8)
=(-6+-sqrt(36+3168))/16
=(-6+-sqrt(3204))/16
=(-6+-6 sqrt(89))/16
=(-3+-3 sqrt(89))/8
Hence:
f(x) = 8(x^2+3/8(1+sqrt(89))x+1)(x^2+3/8(1-sqrt(89))+1)
=1/8(8x^2+3(1+sqrt(89))x+8)(8x^2+3(1-sqrt(89))x+8)
Hence using the quadratic formula twice more:
x = (-3(1+sqrt(89))+-sqrt(9(1+sqrt(89))^2-256))/16
=1/16(-3(1+sqrt(89))+-sqrt(9(90+2sqrt(89))-256))
=1/16(-3(1+sqrt(89))+-sqrt(554+18sqrt(89)))
x = (-3(1-sqrt(89))+-sqrt(9(1-sqrt(89))^2-256))/16
=1/16(-3(1-sqrt(89))+-sqrt(9(90-2sqrt(89))-256))
=1/16(-3(1-sqrt(89))+-sqrt(554-18sqrt(89)))