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

2 Answers
Jun 20, 2017

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

Jun 20, 2017

+-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)