# Bad Solver

### #1

Posted 19 February 2006 - 12:01 PM

solve((x+1)^2=0) --> {x=-1}

but i want to know how many times it happens...!!

i think OS should upgrade this

### #2

Posted 19 February 2006 - 12:32 PM

i think that cp300 solve is very bad because

solve((x+1)^2=0) --> {x=-1}

but i want to know how many times it happens...!!

i think OS should upgrade this

Well the answer is good, no? (-1+1)^2=0^2=0

What do you mean with "how many times it happens"??? You want to konw how many solutions there are? You should just write dim(solve((x+1)^2=0)), as the returned value is a list!

### #3

Posted 20 February 2006 - 03:19 PM

### #4 Guest_pabloec20_*

Posted 09 March 2006 - 04:08 PM

usin os 1 dunno if that has anything to do

### #5

Posted 10 March 2006 - 05:55 PM

Because the original equation is quadratic it has two solutions, but they are both the same. The solution is x=-1, but this is a repeated root. I think most algebra systems will not give you the multiplicity of a root.i did the dim trick and i get "1" what does it mean? the multiplicity is 2 so its not the multiplicity what i am gettin...

usin os 1 dunno if that has anything to do

### #6

Posted 10 March 2006 - 07:16 PM

### #7

Posted 10 March 2006 - 10:43 PM

If a is a root of P(x) with multiplicity m, then a is also a root of dP(x)/dx with multiplicity m-1, d?P(x)/dx? with multiplicity m-2, ...

### #8

Posted 11 March 2006 - 10:31 AM

you can differentiate (x+1)^2, it gives 2(x+1). As -1 is root of (x+1)^2 and of 2(x+1), you can say the multiplicity is at least 2

If a is a root of P(x) with multiplicity m, then a is also a root of dP(x)/dx with multiplicity m-1, d?P(x)/dx? with multiplicity m-2, ...

i know this idea but there is equations that you can not do this,something like (x+a)^2*(x+b )(x+c)

there could be a x=-a answer 3 times.. at last it is to slow to do this

#### 0 user(s) are reading this topic

0 members, 0 guests, 0 anonymous users