Answered: "The complex zeros of f(x)=x4+1 For the function..."

Leroy Little

Answered question

2022-02-04

The complex zeros of

f (x) = x^{4} + 1

For the function

f (x) = x^{4} + 1

:
(a) Factor f into the product of two irreducible quadratics. (Hint: Complete the square by adding and subtracting

2 x^{2}

.)
(b) Find the zeros of f by finding the zeros of each irreducible quadratic.

uporabah5pn

Beginner2022-02-05Added 11 answers

f (x) = x^{4} + 2 x^{2} + 1 - 2 x^{2}

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

Using

(a^{2} - b^{2}) = (a + b) (a - b)

f (x) = (x^{2} + 1 + \sqrt{2 x}) (x^{2} + 1 - \sqrt{2 x}}

Javon Ross

Beginner2022-02-06Added 13 answers

b.
Zeroes

x^{2} + \sqrt{2 x} + 1 = 0, x^{2} - \sqrt{2 x} + 1 = 0

Using quadratic formula which is

x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}

x = \frac{- \sqrt{2} \pm \sqrt{2 - 4 (1) (1)}}{2}, \frac{- \sqrt{2} \pm \sqrt{2 - 4 (1) (1)}}{2}

= \frac{- \sqrt{2} \pm i \sqrt{2}}{2}, \frac{- \sqrt{2} \pm i \sqrt{2}}{2}

No real zeroescomplex zeroes are

\frac{- \sqrt{2} \pm i \sqrt{2}}{2}, \frac{- \sqrt{2} \pm i \sqrt{2}}{2}

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