# Need help to verify the number of solutions

#### idontknow

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).

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#### phillip1882

x is a real number. not an imaginary or complex number.
x^2 +x +L = 0
x = (1 +/- sqrt(1 -4*L))/2

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#### idontknow

For example $$\displaystyle \lambda =1$$ ; $$\displaystyle D=i\sqrt{5}$$ ; $$\displaystyle Re(D)=0$$.

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1 person

#### mathman

Forum Staff
Why is this question repeated?

1 person

#### phillip1882

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

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#### idontknow

I fixed the exponents now: $$\displaystyle N_{\lambda } = \frac{\displaystyle 1 + \lim_{s\rightarrow \infty} \left[-2\left(1+e^{-2s(-1-4\lambda)}\right)^{-1}+1\right]}{2}\lim_{s\rightarrow \infty} \left[-2\left(1+e^{-2s(-1-4\lambda)}\right)^{-1}+1\right]\cdot \left\lceil \frac{Re\{ \sqrt{-1-4\lambda}\}+1}{Re \{ \sqrt{-1-4\lambda}\}+2} \right\rceil.$$

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#### idontknow

Just plug the value of $$\displaystyle \lambda$$ into $$\displaystyle N(\lambda )$$ and it gives the number of solutions .