Question

Which is greater root 7 minus root 3 and root 5 minus root 1?

Answer 1

This is the answer for your question

Answer 2

$i) \sqrt{7} - \sqrt{3} \\= \frac{ ( \sqrt{7} - \sqrt{3})(\sqrt{7} + \sqrt{3})}{(\sqrt{7} + \sqrt{3})} \\= \frac{(\sqrt{7})^{2} - (\sqrt{3})^{2}}{\sqrt{7} + \sqrt{3}} \\= \frac{7 - 3}{\sqrt{7} + \sqrt{3}} \\= \frac{4}{\sqrt{7} + \sqrt{3}} \: ---(1)$

$ii) \sqrt{5} - \sqrt{1} \\= \frac{ ( \sqrt{5} - \sqrt{1})(\sqrt{5} + \sqrt{1})}{(\sqrt{5} + \sqrt{1})} \\= \frac{(\sqrt{5})^{2} - (\sqrt{1})^{2}}{\sqrt{5} + \sqrt{1}} \\= \frac{5 - 1}{\sqrt{5} + \sqrt{1}} \\= \frac{4}{\sqrt{5} + 1} \: ---(2)$

$Now , \sqrt{7} - \sqrt{3} \:\:\:\:\:\: \sqrt{5} - \sqrt{1}$

$\implies \frac{4}{\sqrt{7} + \sqrt{3}} \: \:\:\:\:\:\frac{4}{\sqrt{5} + 1}$

$\implies \frac{4}{\sqrt{7} + \sqrt{3}} \:\pink { <} \: \frac{4}{\sqrt{5} + 1}$

Therefore.,

$(\sqrt{7} - \sqrt{3})\: \pink { < }\: (\sqrt{5} - \sqrt{1})$

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