1Factorise it to (x2+3x+1)(x-1)2
Then d ans is 1,(-3+√5)/2 and (-3-√5)/2
Find the roots of the equation
x4+x3-4x2+x+1=0
Hint: Think of (x+1/x)
-
UP 0 DOWN 0 0 11
11 Answers
1
62Aditya .. this is good work.. but how did you reach the factors?
1We check if factors of the constant term(here 1) satisfies the equation. so x-1 is a factor. here 1 does. Now we write like
x3(x-1)+2x2(x-1)-2x(x-1)-1(x-1)
=(x-1)(x3+2x2-2x-1)
now for x3+2x2-2x-1 we proceed as above and get x-1 again as a factor.
Using the sane steps, x3+2x2-2x-1=(x-1)(x2+3x+1)
So joining them we get the ans as (x-1)2(x2+3x+1)
Now we just find the roots.
62Good work..
I would suggest you to look at one more thing all the time when you have a equation of 4th degree...
Divide the equation by x^2
Try doing that and see if you can think of a solution?
1think i got somthing....
on dividing by x2 we get
x2+x -4 +1/x+1/x2=0
which cab be rearranged as (x-1/x)2 + (x+ 1/x)=2
Now as we know min of x+ 1/x=2 {taking positive values of x,now}
so x=1/x and x>0{we will work for x<0 also}
we get x=1.
1also at x=1,d(x4+x3-4x2+x+1)/dx=0....threfore this is a reapeated root
62no no.. you have to think just a bit more after th erearrangement.. there is a better way to do it...
btw if you din notice.. it was for class IX X :P :)
1now once,we get the 2 roots divide the equation by (x-1)2 and the quotent is a quad. eqn. we can find out its roots[1][1]
62What I meant is this ... divide by x^2 to get...
x2+x-4+1/x+1/x2=0
Thus, x2+1/x2-4+1/x+x=0
Now what should come to your mind is that x+1/x = t
then x2+1/x2=t2-2
Thus, t2-6+t=0
solve for t, then solve for x..
In all you will get 4 roots.. (some of which will be not real)