Alg2.1.17

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Show that $\frac{1}{\sqrt{12-\sqrt{140}}}-\frac{1}{\sqrt{8-\sqrt{60}}}-\frac{2}{\sqrt{10+\sqrt{84}}}=0\,$

Finding the square roots of the denominators separately,

$\sqrt{12-\sqrt{140}}=\sqrt{12-\sqrt{4\cdot35}}\,$

$\sqrt{7+5-2\sqrt{7\cdot5}}\,$

$\sqrt{(\sqrt{7})^{2}+(\sqrt{5})^{2}-2\sqrt{7\cdot5}}\,$

$\sqrt{((\sqrt{7})-(\sqrt{5}))^{2}}\,$

$\sqrt{7}-\sqrt{5}\,$

Square root of the second denominator,

$\sqrt{8-\sqrt{60}}=\sqrt{8-\sqrt{4\cdot15}}\,$

$\sqrt{5+3-2\sqrt{3\cdot5}}\,$

$\sqrt{(\sqrt{5})^{2}+(\sqrt{3})^{2}-2\sqrt{5\cdot3}}\,$

$\sqrt{((\sqrt{5})-(\sqrt{3}))^{2}}\,$

$\sqrt{5}-\sqrt{3}\,$

Square root of the third denominator is

$\sqrt{10+\sqrt{84}}=\sqrt{10+\sqrt{4\cdot21}}\,$

$\sqrt{7+3+2\sqrt{7\cdot3}}\,$

$\sqrt{(\sqrt{7})^{2}+(\sqrt{3})^{2}+2\sqrt{7\cdot3}}\,$

$\sqrt{((\sqrt{7})+(\sqrt{3}))^{2}}\,$

$\sqrt{7}+\sqrt{3}\,$

Now the expression on the lefthand side is

$\frac{1}{\sqrt{7}-\sqrt{5}}-\frac{1}{\sqrt{5}-\sqrt{3}}-\frac{2}{\sqrt{7}+\sqrt{3}}\,$

Rationalising the denominators,we get

$\frac{\sqrt{7}+\sqrt{5}}{2}-\frac{\sqrt{5}+\sqrt{3}}{2}-2\frac{\sqrt{7}-\sqrt{3}}{4}\,$

$\frac{\sqrt{7}+\sqrt{5}-\sqrt{5}-\sqrt{3}-\sqrt{7}+\sqrt{3}}{2}\,$

Simplifying we get the numerator as zero,hence the result is zero which is to be proved.

Main Page:Algebra:Radicals

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