Question

# If a2 – 2a – 1 = 0, then ${a}^{2}+\frac{1}{{a}^{2}}$ = __________.

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Solution

## $\mathrm{Given}:\phantom{\rule{0ex}{0ex}}{a}^{2}-2a-1=0\phantom{\rule{0ex}{0ex}}⇒{a}^{2}-1=2a\phantom{\rule{0ex}{0ex}}\mathrm{Dividing}\text{'}a\text{'}\mathrm{on}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get}\phantom{\rule{0ex}{0ex}}⇒\frac{{a}^{2}-1}{a}=\frac{2a}{a}\phantom{\rule{0ex}{0ex}}⇒\frac{{a}^{2}}{a}-\frac{1}{a}=2\phantom{\rule{0ex}{0ex}}⇒a-\frac{1}{a}=2\phantom{\rule{0ex}{0ex}}\mathrm{Squaring}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get}\phantom{\rule{0ex}{0ex}}⇒{\left(a-\frac{1}{a}\right)}^{2}={2}^{2}\phantom{\rule{0ex}{0ex}}⇒{a}^{2}+{\left(\frac{1}{a}\right)}^{2}-2a\frac{1}{a}=4\left(\mathrm{Using}\mathrm{the}\mathrm{identity}:{\left(a-b\right)}^{2}={a}^{2}+{b}^{2}-2ab\right)\phantom{\rule{0ex}{0ex}}⇒{a}^{2}+\frac{1}{{a}^{2}}-2=4\phantom{\rule{0ex}{0ex}}⇒{a}^{2}+\frac{1}{{a}^{2}}=6$ Hence, if a2 – 2a – 1 = 0, then ${a}^{2}+\frac{1}{{a}^{2}}$ = 6.

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