Respuesta :
Given equation: [tex]\sqrt{2p+1}+2\sqrt{p}=1[/tex].
[tex]\mathrm{Subtract\:}2\sqrt{p}\mathrm{\:from\:both\:sides}[/tex]
[tex]\sqrt{2p+1}+2\sqrt{p}-2\sqrt{p}=1-2\sqrt{p}[/tex]
[tex]\sqrt{2p+1}=1-2\sqrt{p}[/tex]
[tex]\mathrm{Square\:both\:sides}[/tex]
[tex]\left(\sqrt{2p+1}\right)^2=\left(1-2\sqrt{p}\right)^2[/tex]
[tex]2p+1=1-4\sqrt{p}+4p[/tex]
[tex]\mathrm{Subtract\:}1+4p\mathrm{\:from\:both\:sides}[/tex]
[tex]2p+1-\left(1+4p\right)=1-4\sqrt{p}+4p-\left(1+4p\right)[/tex]
[tex]-2p=-4\sqrt{p}[/tex]
[tex]\left(-2p\right)^2=\left(-4\sqrt{p}\right)^2[/tex]
[tex]4p^2=16p[/tex]
[tex]4p^2-16p=0[/tex]
Factoring out 4p, we get
[tex]4p(p-4) =0[/tex]
4p=0 and p-4 =0
p=0 and p=4.
[tex]\mathrm{Check\:the\:solutions\:by\:plugging\:them\:into\:}\sqrt{2p+1}+2\sqrt{p}=1[/tex]
[tex]\mathrm{Plug}\quad p=4:\quad \sqrt{2\cdot \:4+1}+2\sqrt{4}=1\quad \Rightarrow \quad \mathrm{False}[/tex]
[tex]\mathrm{Plug}\quad p=0:\quad \sqrt{2\cdot \:0+1}+2\sqrt{0}=1\quad \Rightarrow \quad \mathrm{True}[/tex]
