What are the zeros of x^4-26x^2+1 ?
2 Answers
Four zeros are
Explanation:
Let
Using quadratic formula zeros of this are given by
or
i.e. either
=
=
=
and
or
=
=
=
and
Hence, four zeros are
Explanation:
"If all you have is a hammer, then everything looks like a nail"
- Abraham Maslow (1966)
Given:
x^4-26x^2+1
It seems to me that the natural tendency is to notice that this is a quadratic in
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...
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))
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)