Help (last one)

If P and V are the two distinct solutions to the equation $x^2=x+1$, then what is the value of (\P-V)^2

$x^2=x+1\\ x^2-x-1=0\\ x=-\frac{p}{2}\pm \sqrt{(\frac{p}{2})^2-q}$

$x=\frac{1}{2}\pm \sqrt{\frac{1}{4}+1}\\ x=\frac{1}{2}\pm \sqrt{\frac{5}{4}}\\ x=\frac{1}{2}\pm \frac{1}{2}\sqrt{5}$

$P=\frac{1}{2}+\frac{1}{2}\sqrt{5}\\ V=\frac{1}{2}-\frac{1}{2}\sqrt{5}$

$(P-V)^2=((\frac{1}{2}+\frac{1}{2}\sqrt{5})-(\frac{1}{2}-\frac{1}{2}\sqrt{5}))^2\\ {\color{red}(P-V)^2=(2\sqrt{5})^2}\ small\ mistake\ \color{blue}(P-V)^2=(\sqrt{5})^2 \ $

$(P-V)^2=5$

Thanks heureka and Allan!

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