Question

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Find the solutions of x⁴+1, then find two distinct square roots of i and -i.
Find the solutions of x²+x+1, then find two distinct square roots of two solutions.

Answer 1

First to find the zeros of x⁴+1, we need to factorize over reals.

→ $x^4+1=(x^2+1)^2-2x^2$

→ $x^4+1=(x^2+\sqrt{2} x+1)(x^2-\sqrt{2} x+1)$

Zeros of two quadratic factors are solutions of x⁴+1.

$\boxed{\text{Quadratic Formula : }x=\frac{-b\pm\sqrt{b^2-4ac} }{2a} }$

$\boxed{\text{Solutions of first factor are }\frac{-\sqrt{2} \pm\sqrt{2}i }{2} }$

$\boxed{\text{Solutions of second factor are }\frac{\sqrt{2} \pm\sqrt{2}i }{2} }$

Meanwhile, two factors of x⁴+1 can be x²+i and x²-i.

Therefore, four zeros of x⁴+1 are square roots of i and -i.

∴$\sqrt{-i} =\frac{-\sqrt{2} \pm\sqrt{2}i }{2}$

∴$\sqrt{i} =\frac{\sqrt{2} \pm\sqrt{2}i }{2}$

To find the zeros of x²+x+1, we use quadratic equations.

$\boxed{\text{Quadratic Formula : }x=\frac{-b\pm\sqrt{b^2-4ac} }{2a} }$

$\boxed{\text{Two solutions are }\frac{-1\pm\sqrt{3}i }{2} }$

If we change place of x with x², we can find out that $x^4+x^2+1=(x^2-\alpha )(x^2-\beta )$. Alpha and beta are two solutions of original quadratic equations. It is required that four solutions of x⁴+x²+1 should be square roots of two solutions.

Meanwhile, x⁴+x²+1 can be factorized over reals.

→ $x^4+x^2+1=(x^2+1)^2-x^2$

→ $x^4+x^2+1=(x^2+x+1)(x^2-x+1)$

$\boxed{\text{Four solutions are }\frac{-1\pm\sqrt{3}i }{2}, \frac{1\pm\sqrt{3}i }{2} }$

$\boxed{\text{Therefore, the square roots are }\frac{-1\pm\sqrt{3}i }{2}, \frac{1\pm\sqrt{3}i }{2} }$