anonymous
  • anonymous
Solve for x . (a^2-1)x^2+2(a-1)x+1=0
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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TuringTest
  • TuringTest
you know you could just use the quadratic formula... but I suppose you want a smarter way?
anonymous
  • anonymous
Smarter way would be perfect.
agreene
  • agreene
completing the square is pretty easy as well. pick your poison... I personally, dont know--and have never used the quadratic formula :)

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TuringTest
  • TuringTest
oh I do \(not\) believe that agreene!
TuringTest
  • TuringTest
but yeah, completing the square sounds a little less painful
anonymous
  • anonymous
And how is it done?
agreene
  • agreene
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.
TuringTest
  • TuringTest
the quadratic formula can be used in all cases
agreene
  • agreene
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: \[x=\frac{\pm i\sqrt{2}-\sqrt{a-1}}{\sqrt{a-1}(a+1)}\]
agreene
  • agreene
and, @TuringTest I thought there were a few cases where it was indeterminate...
anonymous
  • anonymous
What about this. Find the values of a so that the equation (a^2-1)x^2+2(a-1)x+1>0
TuringTest
  • TuringTest
@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
TuringTest
  • TuringTest
(and if a=0 it's not a quadratic obviously)
agreene
  • agreene
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)
TuringTest
  • TuringTest
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
agreene
  • agreene
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
TuringTest
  • TuringTest
it sure can, as one may expect in mathematics they are really different faces of the same thing
agreene
  • agreene
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!

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