What are the zeros of x^4-26x^2+1x4−26x2+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+1x4−26x2+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+13^2-168x4−26x2+1=(x2)2−2(13)x2+132−168
color(white)(x^4-26x^2+1) = (x^2-13)^2-(2sqrt(42))^2x4−26x2+1=(x2−13)2−(2√42)2
color(white)(x^4-26x^2+1) = (x^2-13-2sqrt(42))(x^2-13+2sqrt(42))x4−26x2+1=(x2−13−2√42)(x2−13+2√42)
Hence:
x = +-sqrt(13+-2sqrt(42))x=±√13±2√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(a2−kab+b2)(a2+kab+b2)=a4+(2−k2)a2b2+b4
(a^2-kab-b^2)(a^2+kab-b^2) = a^4-(2+k^2)a^2b^2+b^4(a2−kab−b2)(a2+kab−b2)=a4−(2+k2)a2b2+b4
So we find:
x^4-26x^2+1 = (x^2-2sqrt(7)x+1)(x^2+2sqrt(7)x+1)x4−26x2+1=(x2−2√7x+1)(x2+2√7x+1)
Or:
x^4-26x^2+1 = (x^2-2sqrt(6)x-1)(x^2+2sqrt(6)x-1)x4−26x2+1=(x2−2√6x−1)(x2+2√6x−1)
We can then use our favourite quadratic method to find things like:
x^2-2sqrt(7)x+1 = x^2-2sqrt(7)x+7-6x2−2√7x+1=x2−2√7x+7−6
color(white)(x^2-2sqrt(7)x+1) = (x-sqrt(7))^2-(sqrt(6))^2x2−2√7x+1=(x−√7)2−(√6)2
color(white)(x^2-2sqrt(7)x+1) = (x-sqrt(7)-sqrt(6))(x-sqrt(7)+sqrt(6))x2−2√7x+1=(x−√7−√6)(x−√7+√6)
x^2+2sqrt(7)x+1 = (x+sqrt(7)-sqrt(6))(x+sqrt(7)+sqrt(6))x2+2√7x+1=(x+√7−√6)(x+√7+√6)
Hence the zeros of our quartic are:
+-sqrt(7)+-sqrt(6)±√7±√6