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

 May 23, 2019

Best Answer 

 #1
avatar+8580 
+3

 

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!

laugh  !

 May 23, 2019
edited by asinus  May 23, 2019
 #1
avatar+8580 
+3
Best Answer

 

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!

laugh  !

asinus May 23, 2019
edited by asinus  May 23, 2019

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