solving for zero.. A curvature problem

[SigepBrandon]

\(\displaystyle \frac{\sqrt{x^{2}+1}}{(x^{2}+1)^{3}}(x^{2}+1-3x^{2})\)

I need help solving the above for zero. The original function had a Log, so only positive x's are needed. Actually, I saw an example and know that x comes out to \(\displaystyle \frac{1}{\sqrt{2}}\)...

I tried solving \(\displaystyle \sqrt{x^{2}+1\) and \(\displaystyle {(x^{2}+1)^{3}}\) and \(\displaystyle (x^{2}+1-3x^{2})\) for zero separately, but not a lot of luck... any advice?

 

[Denis]

....\(\displaystyle (x^{2}+1-3x^{2})\) for zero separately, but not a lot of luck... any advice?

Why do you not show that as 1 - 2x^2? Factors to (1 + xSQRT(2))(1 - xSQRT(2))

 

[SigepBrandon]

I left it in that form because of an earlier factoring, but simplified it when setting it equal to zero. I tried to solve it quadratically, but got 1 as an answer (maybe because you can not solve quadratically with b=0?) and did not realize that it factors to (1 + xSQRT(2))(1 - xSQRT(2))

Thanks Denis!

 

[Deleted member 4993]

I left it in that form because of an earlier factoring, but simplified it when setting it equal to zero. I tried to solve it quadratically, but got 1 as an answer (maybe because you can not solve quadratically with b=0?) and did not realize that it factors to (1 + xSQRT(2))(1 - xSQRT(2))

Thanks Denis!

Ax[sup:ta97yvi9]2[/sup:ta97yvi9] + Bx + C = 0

x[sub:ta97yvi9]1,2[/sub:ta97yvi9] = [-B ± ?{B[sup:ta97yvi9]2[/sup:ta97yvi9] - 4AC}]/(2A)

In your case, A = -2, B = 0, C = 1

x[sub:ta97yvi9]1,2[/sub:ta97yvi9] = ± ?{ 8}]/(4) = ± 1/?2

Works everytime!! (except for A = 0, then ofcourse you do not have a quadratic equation)