Solve the equation $\displaystyle x^{4}-4x^{2}+8x+35= 0$ having given that one root is $\displaystyle 2+\sqrt{-3}.$
- The roots of the given equation are $\displaystyle -2\pm i,2\pm\sqrt{3}i.$
- The roots of the given equation are $\displaystyle 2\pm i,2\pm\sqrt{3}i.$
- The roots of the given equation are $\displaystyle \pm2+ i,2\pm\sqrt{3}i.$
- The roots of the given equation are $\displaystyle \pm2- i,2\pm\sqrt{3}i.$
Given root 2 + i*sqrt(3). Since coefficients are real, 2 - i*sqrt(3) is also a root. Sum = 4, product = 4 + 3 = 7. Quadratic factor: x^2 - 4x + 7. Dividing x^4 - 4x^2 + 8x + 35 by x^2 - 4x + 7 gives x^2 + 4x + 5 = 0. Roots are (-4 +/- sqrt(16 - 20))/2 = -2 +/- i. So roots are 2 +/- i*sqrt(3) and -2 +/- i.
Since the coefficients of the equation are real, complex roots must occur in conjugate pairs, making 2 - i sqrt(3) the second root. These two roots come from the quadratic factor x^2 - 4x + 7, which we divide from the original polynomial x^4 - 4x^2 + 8x + 35 to find the other factor. The resulting quadratic factor is x^2 + 4x + 5, and setting it to zero yields the remaining roots using the quadratic formula. Solving x = [-4 +/- sqrt(16 - 20)] / 2 gives -2 +/- i. Therefore, the four roots of the equation are -2 +/- i and 2 +/- sqrt(3)i.