maths
posted by Laila .
the solution set for the following
1.X3+x26x10=0
2.2x24x4=0
3.x4x34x2+x+1=0

1. x^3 + x^2  6x  10 = 0
let f(x) = x^3 + x^2  6x  10
f(1) = ... ≠ 0
f(1) = ... ≠ 0
f(2) = 8 + 4  12  10 ≠ 0
f(2) = 8 + 4 + 12  10 ≠ 0
f(5) ≠0, f(5) ≠ 0
all rational possiblilities have been exhausted, tough from here on in.
Wolfram gave me this ....
http://www.wolframalpha.com/input/?i=x%5E3+%2B+x%5E2++6x++10+%3D+0
x = appr 2.6631, and 2 complex roots
2. I will use completing the square ...
could use the formula, but that would be harder.
first divide by 2
> x^2  2x = 2
x^2  2x + 1 = 2+1
(x1)^2 = 3
x1 = ± √3
x = 1 ± √3
3. let f(x) = x^4  x^3  4x^2 + x + 1 = 0
WOW, even nastier than the first one
Wolfram says this
http://www.wolframalpha.com/input/?i=x%5E4++x%5E3++4x%5E2+%2B+x+%2B+1+%3D+0
What method was suggested to you to solve these?
Have you learned Newton's Method ?