How do you find the complex roots of $a^{4} + a^{2} - 2 = 0$ $a^4+a^2-2=0$ ?

Question

How do you find the complex roots of $a^{4} + a^{2} - 2 = 0$ $a^4+a^2-2=0$ ?

1 Answer

Impact of this question

1443 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-11-24T22:08:48

This quartic has Real roots $a = \pm 1$ $a=+-1$ and Complex roots $a = \pm \sqrt{2} i$ $a=+-sqrt(2)i$

Explanation:

We can write this quartic as a quadratic in $a^{2}$ $a^2$ :

${(a^{2})}^{2} + (a^{2}) - 2 = 0$ $(a^2)^2+(a^2)-2 = 0$

Note that the sum of the coefficients is $0$ $0$ . That is:

$1 + 1 - 2 = 0$ $1+1-2 = 0$

Hence $a^{2} = 1$ $a^2 = 1$ is a solution and $(a^{2} - 1)$ $(a^2-1)$ a factor:

$0 = {(a^{2})}^{2} + (a^{2}) - 2 = (a^{2} - 1) (a^{2} + 2)$ $0 = (a^2)^2+(a^2)-2 = (a^2-1)(a^2+2)$

We can factor this into linear factors with Real or imaginary coefficients using the difference of squares identity:

$A^{2} - B^{2} = (A - B) (A + B)$ $A^2-B^2=(A-B)(A+B)$

as follows:

$(a^{2} - 1) (a^{2} + 2) = (a^{2} - 1^{2}) (a^{2} - {(\sqrt{2} i)}^{2})$ $(a^2-1)(a^2+2) = (a^2-1^2)(a^2-(sqrt(2)i)^2)$

$(a^{2} - 1) (a^{2} + 2) = (a - 1) (a + 1) (a - \sqrt{2} i) (a + \sqrt{2} i)$ $color(white)((a^2-1)(a^2+2)) = (a-1)(a+1)(a-sqrt(2)i)(a+sqrt(2)i)$

So there are Real roots: $a = \pm 1$ $a=+-1$ and Complex roots: $a = \pm \sqrt{2} i$ $a=+-sqrt(2)i$

Science

Math

Humanities

... and beyond

How do you find the complex roots of $a^{4} + a^{2} - 2 = 0$ $a^4+a^2-2=0$ ?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you find the complex roots of a4+a2−2=0a^4+a^2-2=0?

1 Answer

Explanation:

Related questions

Impact of this question

How do you find the complex roots of $a^{4} + a^{2} - 2 = 0$ $a^4+a^2-2=0$ ?