My Math Forum Need help to verify the number of solutions

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 October 29th, 2019, 09:10 AM #1 Senior Member   Joined: Dec 2015 From: Earth Posts: 823 Thanks: 113 Math Focus: Elementary Math Need help to verify the number of solutions Given equation $\displaystyle x^2 +x +\lambda =0 \; ,x\in \mathbb{R}$. Verify N - the number of solutions, using any software (matlab, calculators... etc.) $\displaystyle N_{\lambda } =\frac{\displaystyle 1 + \lim_{s\rightarrow \infty} \left[-2\left(1+e^{-2s(-1-4\lambda)}\right)+1\right]}{2}\lim_{s\rightarrow \infty} \left[-2\left(1+e^{-2s(-1-4\lambda)}\right)+1\right]\cdot \left\lceil \frac{Re\{ \sqrt{-1-4\lambda}\}+1}{Re \{ \sqrt{-1-4\lambda}\}+2} \right\rceil.$ Re - real part of imaginary number (if it exists). Last edited by skipjack; October 30th, 2019 at 02:43 PM.
 October 29th, 2019, 09:29 AM #2 Member   Joined: May 2013 Posts: 57 Thanks: 5 x is a real number. not an imaginary or complex number. x^2 +x +L = 0 x = (1 +/- sqrt(1 -4*L))/2 Last edited by phillip1882; October 29th, 2019 at 10:00 AM.
 October 29th, 2019, 10:58 AM #3 Senior Member   Joined: Dec 2015 From: Earth Posts: 823 Thanks: 113 Math Focus: Elementary Math For example $\displaystyle \lambda =1$ ; $\displaystyle D=i\sqrt{5}$ ; $\displaystyle Re(D)=0$. Thanks from topsquark Last edited by idontknow; October 29th, 2019 at 11:18 AM.
 October 29th, 2019, 01:21 PM #4 Global Moderator   Joined: May 2007 Posts: 6,852 Thanks: 743 Why is this question repeated? Thanks from topsquark
 October 30th, 2019, 12:48 PM #5 Member   Joined: May 2013 Posts: 57 Thanks: 5 it specifically states in the first sentence, x is a real number also, if L = 1; (1 +/- sqrt(1-4*1))/2 (1 +/- sqrt(-3))/2 1+/- i*sqrt(3))/2 x = 1/2 +i*sqrt(3)/2 or x = 1/2 -i*sqrt(3)/2 so the real part is 1/2. and will be 1/2 for any L value > 1/4 Thanks from idontknow Last edited by phillip1882; October 30th, 2019 at 12:52 PM.
 October 30th, 2019, 01:10 PM #6 Senior Member   Joined: Dec 2015 From: Earth Posts: 823 Thanks: 113 Math Focus: Elementary Math I fixed the exponents now: $\displaystyle N_{\lambda } = \frac{\displaystyle 1 + \lim_{s\rightarrow \infty} \left[-2\left(1+e^{-2s(-1-4\lambda)}\right)^{-1}+1\right]}{2}\lim_{s\rightarrow \infty} \left[-2\left(1+e^{-2s(-1-4\lambda)}\right)^{-1}+1\right]\cdot \left\lceil \frac{Re\{ \sqrt{-1-4\lambda}\}+1}{Re \{ \sqrt{-1-4\lambda}\}+2} \right\rceil.$ Last edited by skipjack; October 30th, 2019 at 02:47 PM.
 November 2nd, 2019, 07:54 AM #7 Senior Member   Joined: Dec 2015 From: Earth Posts: 823 Thanks: 113 Math Focus: Elementary Math Just plug the value of $\displaystyle \lambda$ into $\displaystyle N(\lambda )$ and it gives the number of solutions .

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