Let $\lambda_1=\frac{1-\sqrt{5}}{2}$ and $\lambda_2=\frac{1+\sqrt{5}}{2}$.
Prove that $\lambda_i^2=1+\lambda_i$ for $i=1,2$.
Attempt:
$\lambda_1^2=\frac{1-5}{4}=\frac{6-2\sqrt{5}}{4}$.
I don't see how this get's me anywhere. How do I proceed?
Prove that $\lambda_i^2=1+\lambda_i$ for $i=1,2.$
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algebra-precalculus
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0$\lambda_1^2=\cfrac{\left(1-\sqrt{5}\right)^2}{4} \ne -1\,$ – 2017-01-24
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0@dxiv Oh, of course. It is late! – 2017-01-24
2 Answers
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$$\lambda_1^2=\left(\frac{1-\sqrt{5}}{2}\right)^2=\frac{6-2\sqrt{5}}{4}=\frac{3-\sqrt{5}}{2}$$
And $$1+\lambda_1=\frac{1-\sqrt{5}}{2}+1=\frac{1-\sqrt{5}+2}{2}=\frac{3-\sqrt{5}}{2}$$
For $\lambda_2$ do the same.
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Hint: calculate $a=\lambda_1+\lambda_2$ and $b=\lambda_1\,\lambda_2\,$, then $\lambda_{1,2}$ must be the roots of $x^2-a x + b = 0\,$.