Question

What are the zeros of `x^4-26x^2+1` ?

Answer 1

Four zeros are `sqrt7+sqrt6`, `-sqrt7-sqrt6`, `sqrt7-sqrt6` or `sqrt6-sqrt7`

Explanation:

Let `x^2=t`,, then we can write `x^4-26x^2+1` as

`t^2-26t+1`

Using quadratic formula zeros of this are given by

`(26+-sqrt(26^2-4xx1xx1))/2=(26+-sqrt672)/2`

or `13+-2sqrt42`

i.e. either `x^2=13+2sqrt42`

= `7+6+2sqrt42`

= `(sqrt7)^2+(sqrt6)^2+2xxsqrt7xxsqrt6`

= `(sqrt7+sqrt6)^2`

and `x=sqrt7+sqrt6` or `-sqrt7-sqrt6` ................(A)

or `x^2=13-2sqrt42`

= `7+6-2sqrt42`

= `(sqrt7)^2+(sqrt6)^2-2xxsqrt7xxsqrt6`

= `(sqrt7-sqrt6)^2`

and `x=sqrt7-sqrt6` or `sqrt6-sqrt7` ................(B)

Hence, four zeros are `sqrt7+sqrt6`, `-sqrt7-sqrt6`, `sqrt7-sqrt6` or `sqrt6-sqrt7`

Answer 2

`+-sqrt(7)+-sqrt(6)`

Explanation:

"If all you have is a hammer, then everything looks like a nail"
- Abraham Maslow (1966)

`color(white)()`
Given:

`x^4-26x^2+1`

It seems to me that the natural tendency is to notice that this is a quadratic in `x^2`.

Let's see what happens if we just follow that course (using completing the square, though the quadratic formula is basically the same)...

`x^4-26x^2+1 = (x^2)^2-2(13)x^2+13^2-168`

`color(white)(x^4-26x^2+1) = (x^2-13)^2-(2sqrt(42))^2`

`color(white)(x^4-26x^2+1) = (x^2-13-2sqrt(42))(x^2-13+2sqrt(42))`

Hence:

`x = +-sqrt(13+-2sqrt(42))`

We can then attempt to simplify this. This is a perfectly viable approach, but it may be easier to do something a little different...

`color(white)()`
Alternatively, we can use one of the following factorisations:

`(a^2-kab+b^2)(a^2+kab+b^2) = a^4+(2-k^2)a^2b^2+b^4`

`(a^2-kab-b^2)(a^2+kab-b^2) = a^4-(2+k^2)a^2b^2+b^4`

So we find:

`x^4-26x^2+1 = (x^2-2sqrt(7)x+1)(x^2+2sqrt(7)x+1)`

Or:

`x^4-26x^2+1 = (x^2-2sqrt(6)x-1)(x^2+2sqrt(6)x-1)`

We can then use our favourite quadratic method to find things like:

`x^2-2sqrt(7)x+1 = x^2-2sqrt(7)x+7-6`

`color(white)(x^2-2sqrt(7)x+1) = (x-sqrt(7))^2-(sqrt(6))^2`

`color(white)(x^2-2sqrt(7)x+1) = (x-sqrt(7)-sqrt(6))(x-sqrt(7)+sqrt(6))`

`color(white)()`

`x^2+2sqrt(7)x+1 = (x+sqrt(7)-sqrt(6))(x+sqrt(7)+sqrt(6))`

Hence the zeros of our quartic are:

`+-sqrt(7)+-sqrt(6)`