MHB How Do You Solve the Equation \(\sqrt{x+8} + 2 = \sqrt{x}\) in Complex Numbers?

[Amer]

how to solve this equation in complex

\sqrt{x+8} + 2 = \sqrt{x}

Thanks

 

[Sudharaka]

Re: Complex solve a equation

how to solve this equation in complex

\sqrt{x+8} + 2 = \sqrt{x}

Thanks

Hi Amer, :)

Suppose that a complex solution exists for this equation. Let,

\[\sqrt{x+8}=r_1 e^{i\theta_1}\mbox{ and }\sqrt{x}=r_2 e^{i\theta_2}\]

Then by \(\sqrt{x+8} + 2 = \sqrt{x}\) we get,

\[r_1\cos\theta_1+2=r_2\cos\theta_2~~~~~~~~~~~(1)\]

and,

\[r_1\sin\theta_1=r_2\sin\theta_2~~~~~~~~~~~(2)\]

Also,

\[\sqrt{x+8}=r_1 e^{i\theta_1}\mbox{ and }\sqrt{x}=r_2 e^{i\theta_2}\]

\[\Rightarrow r^{2}_2 e^{2i\theta_2}+8=r^{2}_1 e^{2i\theta_1}\]

\[\Rightarrow r^{2}_1\sin2\theta_1=r^{2}_2\sin2\theta_2\]

\[\Rightarrow r^{2}_1\sin\theta_1\cos\theta_1=r^{2}_2\sin\theta_2\cos\theta_2~~~~~~~~~~~(3)\]

By (2) and (3),

\[r_1\cos\theta_1=r_2\cos\theta_2~~~~~~~~~~~(4)\]

Therefore by (1) and (4),

\[2=0\]

which is a contradiction. Hence there exist no solutions for the equation,

\[\sqrt{x+8} + 2 = \sqrt{x}\]

Wolfram verifies this. :)

Kind Regards,
Sudharaka.

 

[chisigma]

Re: Complex solve a equation

how to solve this equation in complex

\sqrt{x+8} + 2 = \sqrt{x}

Thanks

It is fully evident that x=1 [which is a complex number with imaginary part equal to zero...] is solution of the equation if one takes the negative square root in both terms. This solution can easily be found writing the equation as... $\displaystyle \sqrt{x+8} - \sqrt{x} = -2$ (1)... and the applying the standard 'double squaring' procedure... Kind regards $\chi$ $\sigma$