Question

SOLVE EQUATION.
(X^4)-(4x^3)+(3x^2)+8x-16=0

Answer 1

x^40?

Answer 2

Oh its (x^4) sorry

Answer 3

by definition, you are going to substitute x= 1,-1, -2,2,4,-4, 8,-8, and 16,-16. I have tried upto 4 and -4. None of them worked. so substitute other numbers and see if the equation holds true.

Answer 4

And then you are going to perform long division; divide the original eqn by (x-(the number that held eqn after substitution), and after long division, you are going to simply have quadratic equation. THen you can perform familiar quad formula or factoring to get the other roots. All three are possible answers for the system.

Answer 5

Oh yeah I tried that but couldn't find a substitution # for x because none of them worked:(

Answer 6

In that case, there is no clear INTEGER solution

Answer 7

di u find any answer to the equation?

Answer 8

If you couldnt find any integer solution, it is probably sufficient to answer that there is no solution. Unless you are in university, taking at least second year linear algebra course.

Answer 9

In which case, we can talk about a theorem that gives us noninteger solutions. (it is sort of similar to quadratic formula in essence).

Answer 10

I thought with all quartic equations, the roots would always be
all 4 real OR
all 4 complex (imaginary) OR
2 real roots and 2 complex (imaginary) roots.
I don't think it ever winds up with just 1 real root.

Answer 11

You'll get two quadratic factors, one with 2 reals, the other with 2 imaginary. Here are the factors: (x^2-3 x+4) (x^2-x-4) = 0

Answer 12

Gee, I don't know how you factored that but the right hand equation contains two real roots.
x² -x -4 =0
The quadratic formula gives these results:
2.56155281280883
-1.56155281280883

Notice that the decimal portion of both roots is exactly the same?
Well, anyway there's 2 real roots

Answer 13

Here's one way to factor the quartic:
http://www.sosmath.com/algebra/factor/fac12/fac12.html
I found wolfram a lot more convenient though.

Answer 14

If nothing else, factoring that quartic really brings this problem to a conclusion.

As for the complex (imaginary) roots solving for x^2-3 x+4:
x= [-(-3) ± sqrt(9 - 4*1*4)] / 2*1
x = 3 ±sqrt(9 -16) / 2
x = (3 ±sqrt(-7)) / 2

x = 1.5 plus sqrt(-7)/2
x = 1.5 minus sqrt(-7)/2