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pretend X^4 as x^2
Essentially what louis said. Make a subsitution u=x^2
so its like y^2-10y^2+9=0
then when u get some thing like (y-21551515)(y-611919) substitute x^2 back in
(BTW JUST made up the numbers.) i didnt solve it
i still dont undertsand..i can do the factoring for that equation but where does theother x^2 come in?
the factored form of that equation would be (y-9)(y-1) and then i do?
x squared and x squared, 2x and 5x, 1 and 2, and 3 and 3
x^4-10x^2+9=0 let u = x^2 and if we square u: u^2 = x^4 replace x^4 with u^2 and x^2 with u u^2 - 10u +9 =0 solve for u once you get u, solve for x: remember u= x^2 so x= ± sqrt(u)
I just factored this out, and not everything is able to be factorable @sanguinepenguin95
(y-9)(y-1) = 0 that means either (y-9) is zero or (y-1) is zero. you get 2 equations y-9 =0 y-1=0
yo u guys are making it too hard for him.... loook once u get (y-9)(y-1)... put X^2 into the "y"s and solve
There are two monomials are not able to be factored @sanguinepenguin95
It is not able to be factored any more, because there are two monomials that can't be factored any further @sanguinepenguin95
so it would be (x+3)(x-3)(x+1)(x-1)?
then to check plug the answers back in the orignial equations and whatever doesnt equals ... u cross off that answers. they're called like extraneous values i believe
Thankyou so much :)
x=3 -3 1 -1.... then plug them in the equation and whatever doesnt equals u cross them off
yeah no problem
I don't think I would do it quite that way. after you find y= 9 and y=1 you use x = ± sq rt(y) to get x= ± sqrt(9) and x= ± sqrt(1) in other words: x= +3, x= -3, x= +1 and x= -1
you have a 4th degree polynomial, so you should expect 4 (possibly repeated) roots.
... umm this is like alg 2 so id think theirs going to be 4th degree polynomial for this type of question..... maybe in calc or w/e
Actually, this problem is very typical of an algebra II course.