- #1

- 15

- 0

^{4}+2x

^{2}+9

If i do it by quadratic formula, i get complex roots ... in my book, it has been factorized to

(x

^{2}+2x+3)(x

^{2}-2x + 3)

- Thread starter sarah786
- Start date

- #1

- 15

- 0

If i do it by quadratic formula, i get complex roots ... in my book, it has been factorized to

(x

- #2

I like Serena

Homework Helper

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Which complex roots did you get?^{4}+2x^{2}+9

If i do it by quadratic formula, i get complex roots ... in my book, it has been factorized to

(x^{2}+2x+3)(x^{2}-2x + 3)

- #3

Mentallic

Homework Helper

- 3,798

- 94

When it says factorize over the reals, it means your factors have to have real coefficients. A root of x-(a+bi) does not have all real coefficients. Also remember that any real quadratic with a complex root also has a complex conjugate as its other root. So you can also work backwards from the two conjugates to find the real quadratic that has those roots.^{4}+2x^{2}+9

If i do it by quadratic formula, i get complex roots ... in my book, it has been factorized to

(x^{2}+2x+3)(x^{2}-2x + 3)

- #4

- 118

- 1

your equation can be written as

X

= X

= ( X

= ( X

Last edited:

- #5

mathman

Science Advisor

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Let u=x^{4}+2x^{2}+9

If i do it by quadratic formula, i get complex roots ... in my book, it has been factorized to

(x^{2}+2x+3)(x^{2}-2x + 3)

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