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Year 11 Maths - Specialist Geometry in the plane and proof

The golden ratio

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Questions
Question 1
75499

Given that the golden ratio \(\phi = \dfrac{a}{b} = \dfrac{{a + b}}{a}\), find the exact value of \(\phi\)

\(\phi=\dfrac{1+\sqrt{5}}{2}\)


$$
\begin{aligned}
\frac{a}{b}&=\frac{a+b}{a} \\
\frac{a}{b}&=\frac{a}{a}+\frac{b}{a} \\
\phi&=1+\frac{1}{\phi}\\
\phi^{2}&=\phi+1\\
\phi^{2}-\phi-1&=0\\
\phi &=\frac{1 \pm \sqrt{1-4 \times 1\times-1}}{2\times 1}\\
&=\frac{1 \pm \sqrt{5}}{2} \\
\therefore\ \phi &=\frac{1+\sqrt{5}}{2}
\end{aligned}
$$

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