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Question Number 59600 by Gulay last updated on 12/May/19

Answered by MJS last updated on 12/May/19

$${(a+b\sqrt{3})}^2=14+3\sqrt{3}$$$$a^2+3b^2+2ab\sqrt{3}=14+3\sqrt{3}$$$$\Rightarrow$$$$a^2+3b^2=14$$$$2ab=3\Rightarrow b=\frac{3}{2a}$$$$a^2+\frac{27}{4a^2}=14$$$$a^4-14a^2+\frac{27}{4}=0\Rightarrow a^2=\frac{1}{2}\vee a^2=\frac{27}{2}$$$$\Rightarrow a=\pm \frac{\sqrt{2}}{2}\vee a=\pm \frac{3\sqrt{6}}{2}$$$$\Rightarrow b=\pm \frac{3\sqrt{2}}{2}\vee b=\pm \frac{\sqrt{6}}{6}$$$$\Rightarrow \sqrt{14+3\sqrt{3}}=\frac{\sqrt{2}}{2}(1+3\sqrt{3})$$$$\Rightarrow \sqrt{14-3\sqrt{3}}=\frac{\sqrt{2}}{2}(-1+3\sqrt{3})$$$$\Rightarrow \sqrt{14+3\sqrt{3}}-\sqrt{14-3\sqrt{3}}=\sqrt{2}$$

Commented by Kunal12588 last updated on 12/May/19

$$sirwhy\sqrt{14-3\sqrt{3}}\ne \frac{\sqrt{2}}{2}(1-3\sqrt{3})$$$$isitbcoz\frac{\sqrt{2}}{2}(1-3\sqrt{3})<0?$$

Commented by MJS last updated on 12/May/19

$$\sqrt{x}>0\mathrm{for}x>0$$$$\sqrt[n]{x}\mathrm{is}\mathrm{always}\mathrm{unique}\mathrm{for}x\in \mathbb{C}$$$${\mathrm{it}}^{\prime }\mathrm{s}\mathrm{not}\mathrm{an}\mathrm{equation}!$$$$z=r{\mathrm{e}}^{\mathrm{i}\theta }$$$$\sqrt[n]{z}=\sqrt[n]{r}{\mathrm{e}}^{\mathrm{i}\frac{\theta }{n}}$$

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