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Question Number 97479 by Hassen_Timol last updated on 08/Jun/20

Commented by Hassen_Timol last updated on 08/Jun/20

$$\mathrm{How}\mathrm{can}\mathrm{I}\mathrm{prove}\mathrm{these}\mathrm{equalities}\mathrm{please}?$$

Commented by john santu last updated on 08/Jun/20

$$\mathrm{in}\mathrm{generally}$$$$\sqrt{\mathrm{a}+\mathrm{b}+2\sqrt{\mathrm{ab}}}=\sqrt{\mathrm{a}}+\sqrt{\mathrm{b}}$$$$\sqrt{\mathrm{a}+\mathrm{b}-2\sqrt{\mathrm{ab}}}=\sqrt{\mathrm{a}}-\sqrt{\mathrm{b}};\mathrm{a}>\mathrm{b}$$

Answered by Aziztisffola last updated on 08/Jun/20

$$4+2\sqrt{3}=1+2\times 1\times \sqrt{3}+{\left(\sqrt{3}\right)}^2={(1+\sqrt{3})}^2$$$$\sqrt{4+2\sqrt{3}}=\sqrt{{(1+\sqrt{3})}^2}=1+\sqrt{3}$$

Commented by Hassen_Timol last updated on 08/Jun/20

$$\mathrm{Thanks}\mathrm{a}\mathrm{lot}.\mathrm{I}{\mathrm{didn}}^{\prime }\mathrm{t}\mathrm{think}\mathrm{about}\mathrm{that}...$$

Answered by Aziztisffola last updated on 08/Jun/20

$${(\sqrt{3+\sqrt{5}}-\sqrt{3-\sqrt{5}})}^2$$$$=3+\sqrt{5}+3-\sqrt{5}-2\sqrt{(3+\sqrt{5})(3-\sqrt{5})}$$$$=6-2\sqrt{9-5}=6-2\times 2=2$$$$\sqrt{3+\sqrt{5}}-\sqrt{3-\sqrt{5}}=\sqrt{2}$$$$$$

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