Question

solve the equation \(2x^4+3x^3-4x^2-3x+2=0\)

Answer 1

first i what i did is below

2\((x^4+1\))+3\((x^3-x\))-\(4x^2\)=0

Answer 2

multiplying both sides by \(\dfrac1x^2\)

2\((x^2+\dfrac{1}{x^2}\))+3\((x-\dfrac{1}{x}\))-4=0

Next what we will do

Answer 3

@phi @math&ing001 @Mokeira @midhun.madhu1987 @fabiomartins @Abhisar @hero @helper99

Answer 4

http://www.geteasysolution.com/2x%5E4-3x%5E2-2=0

Answer 5

i did google ..but i don't understand @Mokeira

Answer 6

@midhun.madhu1987 @sasogeek

Answer 7

neither do I...let me try it though and see if i can help you

Answer 8

@yayanieves @Jaynator495 @StarBreakerMegg @cj49

Answer 9

@MilenaSaeger

Answer 10

@terenzreignz

Answer 11

im sorry but I cant help

Answer 12

Try Rational Root test.

Answer 13

@bibby @undeadknight26

Answer 14

I usually try to spot patterns in questions like this. For example in your equation:\[2x^4+3x^3-4x^2-3x+2=0\]I would first try to factor the \(x^4\) and \(x^2\) terms together and then the \(x^3\) and \(x\) terms to get:\[2x^2(x^2-2)+3x(x^2-1)+2=0\]Now I can see that I should try and make both the first 2 terms have \((x^2-1)\) so that I can factor it out. So I woud proceed as follows:\[2x^2(x^2-1)-2x^2+3x(x^2-1)+2=0\]Hopefully you can continue from here...

Answer 15

no @asnaseer i am dong not in this way not by factorization by taking other way

Answer 16

@ziqbal103 - please explain what you mean - I don't quite understand what you are trying to say?

Answer 17

Ohh... right.

dangit, asnaseer... I really hate finding roots to long polynomials XD

Answer 18

x=-1,-2,1

Answer 19

https://www.youtube.com/watch?v=xY7Vpxvm62A

Answer 20

great link

Answer 21

@triciaal @taylor12344 @Luigi0210 @liliy

Answer 22

@kirbykirby

Answer 23

did you check the youtube link cj posted?

Answer 24

yes but i prepare these uestion not in this way that @cj49 sent the link...@zimmah

Answer 25

well, i wouldn't know how to do it any other way

Answer 26

except calculator

Answer 27

i am going to copying
(\(x+\dfrac{1}{x}\))=t , then \((x^2+\dfrac{1}{x^2}-2\))=\(t^2\) or \(x^2+\dfrac{1}{x^2}\)=\(t^2+2\)

Answer 28

in this method i want to solve my question

Answer 29

in that case you would be better off using:\[t=x-\frac{1}{x}\]

Answer 30

yes do u know this method @asnaseer

Answer 31

which also gives:\[t^2+2=x^2+\frac{1}{x^2}\]

Answer 32

@ziqbal103 - no I have never seen this method. Does it have a name?

Answer 33

equation in exponenetial form @asnaseer

Answer 34

if you use the substitutions I gave above, your equation reduces to:\[2(t^2+2)+3t-4=0\]which is a simple quadratic.

Answer 35

yes but in above i don't know y they take square root of \(x-\dfrac1{x}\)

Answer 36

what do you mean?

Answer 37

2x4+3x3−4x2−3x+2=0
I don't have time to look at all the responses
one approach is to apply the remainder theorem to start and then factorize
if the value of x is a factor then the remainder = 0
2 * 2 = 4
factors of 4 1, -1. 2, -2
f(1) = 2 + 3 - 4 - 3 +2 = 0 ; x=1 (x-1) is a factor
f(-1)= 2 - 3 -4 + 3 + 2 = 0; x = -1 (x + 1) is a factor
f(2) =
f(-2)=
the degree of polynomial and the rule of signs will let you know how many roots you can have in this case maximum of 4

Answer 38

@ziqbal103 - you ended up with:\[2(x^2+\frac{1}{x^2})+3(x-\frac{1}{x})-4=0\]if we use the substitution:\[t=x-\frac{1}{x}\implies t^2=x^2+\frac{1}{x^2}-2\]we get:\[2(t^2+2)+3t-4=0\]\[\therefore 2t^2+4+3t-4=0\]\[\therefore 2t^2+3t=0\]\[\therefore t(2t+3)=0\]\[\therefore t=0\text{ or }-\frac{3}{2}\]so next you would use:\[t=x-\frac{1}{x}\]to solve for \(x\) for each value of \(t\)

Answer 39

repost of @asnaseer
which also gives:
t2+2=x2+1x2
solve this quadratic will give t = 0 , t = -3/2
x = 1, -1, 1/2, -2

Answer 40

in this question
6x^4-35x^3+62x^2-35x+6

in this multiplying by both sides by \(\dfrac1{x^2}\) we get
6\((x^2+\dfrac1{x^2}\))-35\((x-\dfrac1{x}\))+62=0

let \(x-\dfrac1{x}\)=t ,then \(x^2+\dfrac1{x^2}+2\)=\(t^2\)

in above why we don't take the square of \(x-\dfrac1{x}\)

Answer 41

@asnaseer

Answer 42

@goformit100 please halp me

Answer 43

\[(x-\frac{1}{x})^2=x^2+\frac{1}{x^2}-2\]so we ARE squaring it

Answer 44

but in my guide the procedure is diff. and i don't understand..i solved it but now i forget @asnaseer

Answer 45

I am finding it very difficult to understand where you are getting stuck.
What is this "guide" that you are following? Is it some text book?

Answer 46

yes it is text book @asnaseer

Answer 47

It might help if you uploaded a picture of the page describing the procedure that you are following

Answer 48

it's a pretty complicated method they use in that textbook

Answer 49

what is the name of this text book?