Question

Solve for x . (a^2-1)x^2+2(a-1)x+1=0

Answer 1

you know you could just use the quadratic formula...
but I suppose you want a smarter way?

Answer 2

Smarter way would be perfect.

Answer 3

completing the square is pretty easy as well. pick your poison... I personally, dont know--and have never used the quadratic formula :)

Answer 4

oh I do \(not\) believe that agreene!

Answer 5

but yeah, completing the square sounds a little less painful

Answer 6

And how is it done?

Answer 7

I think my algebra II teacher in gradeschool tried to teach me the quadratic, but i already knew completing the square, and it--unlike the quadratic--can be used in all cases, and all polynomials.

Answer 8

the quadratic formula can be used in all cases

Answer 9

divide by a^2-1
\[\frac{2(a-1)x}{a^2-1}+\frac{1}{a^2-1}+x^2=0\]
subtract 1/(a^2-1)
\[\frac{2(a-1)x}{a^2-1}+x^2=-\frac{1}{a^2-1}\]
add (a-1)^2/(a^2-1)^2 to both sides
\[\frac{2(a-1)x}{a^2-1}+\frac{(a-1)^2}{(a^2-1)^2}+x^2=\frac{(a-1)^2}{(a^2-1)^2}-\frac{1}{a^2-1}\]
factor
\[(\frac{a-1}{a^2-1}+x)^2=-\frac{2}{(a-1)(a+1)^2}\]
sqrt both sides
\[|\frac{a-1}{a^2-1}+x|=-\frac{i\sqrt{2}}{\sqrt{a-1}(a+1)}\]
from there--break the abs and do some simplifications you will find:
\[x=\frac{\pm i\sqrt{2}-\sqrt{a-1}}{\sqrt{a-1}(a+1)}\]

Answer 10

and, @TuringTest I thought there were a few cases where it was indeterminate...

Answer 11

What about this. Find the values of a so that the equation (a^2-1)x^2+2(a-1)x+1>0

Answer 12

@agreene I'd like to see one
the formula is\[ax^2+bx+c=0\implies x={-b\pm\sqrt{b^2-4ac}\over2a}\]so unless a=0 this is defined
if b^2-4ac<0 you get a complex answer, which is still defined

Answer 13

(and if a=0 it's not a quadratic obviously)

Answer 14

yeah, it seems like it would work for all quadratics. I seem to remember someone telling me there were large limitations on it (aside from it only working on quadratics)

Answer 15

I think they misrepresented the situation
the QF always works, but it is sort of a default option in my opinion, as it leads to messy algebra at times like these

Answer 16

yeah, completing the square has the drawback of needing to find sometimes odd algebra to force the factorization... QF looks like it could just be a derived version of CTS

Answer 17

it sure can, as one may expect in mathematics
they are really different faces of the same thing

Answer 18

yeah, if u solve ax^2+bx+c=0 with respect to x, using completing the square... the answer is the quadratic... great... now i know the quadratic and how to get it :( I DIDNT WANT TO KNOW THAT TURING!