Question

the roots of the equation x^5-4x^2+8x+35=0 given that 2+root 3i is a root​

Answer 1

Answer:

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Answer 2

Answer:

Since 2+√3 is a root and the coefficients are real, the second root has to be 2-i√3

Thus, (x - 2 - i√3) (x - 2 + i√3) - x^2 - 2x + ix√3 - 2x + 4 - i2√3 - ix√3 + i2√3 + 3

= x^2 - 4x + 7 is a root of the given equation.

Dividing the original equation by x^2 - 4x + 7 we have (x^4 - 4x^2 + 8x + 35) /(x^2 - 4x + 7) - x^2 + 4x + 5

Thus, we obtain the other root as x^2 + 4x + 5

Again factorizing this root, we get x-

$\frac{ - 4 + \sqrt{16 - 20} }{2}$

=

$\frac{ - 4 + 2i}{2}$

=

$- 2 + i$

The four roots are therefore 2 + i√3, 2 - i√3, -2 + i, -2 - i