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# Help (last one)

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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

#1
+9778
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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-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!

!

May 23, 2019
edited by asinus  May 23, 2019

#1
+9778
+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!

!

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