Posted by **Laila** on Tuesday, October 8, 2013 at 1:29pm.

the solution set for the following

1.X3+x2-6x-10=0

2.2x2-4x-4=0

3.x4-x3-4x2+x+1=0

- maths -
**Reiny**, Tuesday, October 8, 2013 at 2:51pm
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

(x-1)^2 = 3

x-1 = ± √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 ?

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