Question

The number of distinct real roots for x^4-4x^3+12x^2+x-1=0

Answer 1

First of all, it has 4 roots. Firstly, look at the number of sign changes. In this case, it has 3 changes, so it can have 3 or 1 possible positive real roots.

Now substitute x for -x, and look at the signs again: x^4+4x^3+12x^2-x-1=0. It has one change, so it has one negative real root. So finally, the chances are:

3 positive, 1 negative, 0 complex
or
1 positive, 1 negative, 2 complex

Just find out which is the right one.

Answer 2

the graph gives this 1 positive 1 negative and 2 complex
0.31
-0.26

Answer 3

Thats Wat a Real Problems is Using Descrates Sign's Rule We Find 3 positive, 1 negative, 0 complex
or
1 positive, 1 negative, 2 complex

So How to Find the Exact Number...i tried out Factorizing but went in Vain..

Answer 4

@electrokid

Answer 5

f(0) = ?
f(1)=?

only the signs.. no need for exact values
if they are opposite signs, there must be a root in there.

Answer 6

the ratio test fails to find the roots given by @badhi

Answer 7

following @Kikazo's method, you get the two roots and by trial-error method, you can estimate their intervals

Answer 8

hint 2) use the second derivative....
when the f(x) crosses x-axis, so does f''(x)

Answer 9

but the second one does not work either.. that gives you two complex roots.
so, you end up with two real roots!

Answer 10

ur answer is correct but sorry i dont get u

Answer 11

@electrokid

Answer 12

@Kikazo explained it much better.

Answer 13

did you understand that?

Answer 14

Yes

Answer 15

But that only Gives the Idea Abt Possible roots @electrokid

Answer 16

yes. then you can use the first derivative to find the critical points.. the possible maxima and/or minima
the second derivative tells you how many maxima and minima you've got.
case 1) only one maxima (>0) or a minima(<0) -> two possible roots
case 2) one maxima (>0) and one minima (<0) -> 3 possible roots
and so on. and I mena "real" roots.

Answer 17

That Makes sense thxxx

Answer 18

yw.