Author Topic: exact solutions (Read 656 times) Tweet Share

naved_s9994 · « **on:** May 01, 2010, 02:01:53 pm »

find the exact solutions of
x^4 - x^3 + x - 1 = 0

superflya · « **Reply #1 on:** May 01, 2010, 02:15:05 pm »

1 is ur linear factor

divide all by x-1, this yields x^3 +1.

so ur solutions would be x=1 and x=-1

naved_s9994 · « **Reply #2 on:** May 01, 2010, 02:16:22 pm »

Quote from: superflya on May 01, 2010, 02:15:05 pm

1 is ur linear factor

divide all by x-1, this yields x^3 +1.

so ur solutions would be x=1 and x=-1

ahhhhhhhh!!!
Thanks

- Man not sure, WHY I couldnt pick the question up so cleanly!?

Thanks anyways

m@tty · « **Reply #3 on:** May 01, 2010, 02:17:48 pm »

Well immediately you should recognise that 1 is a solution, and -1.

$(1)^4-(1)^3+1-1=0$

and

$(-1)^4-(-1)^3+(-1)-1=0$ .

So you know that two factors are $(x+1)(x-1)$ , or $x^2-1$

Let $(x^2-1)(ax^2+bx+c)=x^4-x^3+x-1$

$(x^2-1)(x^2-x+1)=x^4-x^2-x^3+x+x^2-1=x^4-x^3+x-1$

So the other solutions will be that of $x^2-x+1=0$

Which, using the quadratic formula, are:

$x=\frac{1\pm \sqrt{1-4}}{2}$

Which has no real solutions.

So the only real solutions to that equation are -1, 1.

the.watchman · « **Reply #4 on:** May 01, 2010, 02:42:38 pm »

Or simple grouping method:

$x^4 - x^3 + x -1 =0$

$x^3(x -1) + (x-1)=0$

$(x^3 +1)(x-1)=0$

$x=-1,1$

naved_s9994 · « **Reply #5 on:** May 01, 2010, 02:50:04 pm »

Thanks guys,
Not sure why I couldnt get it really..
So simple!

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Author Topic: exact solutions (Read 656 times) Tweet Share

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