sqrt{10 - sqrt{24} - sqrt{40} + sqrt{60}} = ? | vedclass

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Question

(B) Let $10 - \sqrt{24} - \sqrt{40} + \sqrt{60} = (\sqrt{a} + \sqrt{b} - \sqrt{c})^2$. Expanding the right side: $a + b + c + 2\sqrt{ab} - 2\sqrt{bc}

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Let $10 - \sqrt{24} - \sqrt{40} + \sqrt{60} = (\sqrt{a} + \sqrt{b} - \sqrt{c})^2$. Expanding the right side: $a + b + c + 2\sqrt{ab} - 2\sqrt{bc} - 2\sqrt{ca}$. Comparing terms: $a + b + c = 10$,$2\sqrt{ab} = \sqrt{60} = 2\sqrt{15} \implies ab = 15$,$2\sqrt{bc} = \sqrt{40} = 2\sqrt{10} \implies bc = 10$,$2\sqrt{ca} = \sqrt{24} = 2\sqrt{6} \implies ca = 6$. Multiplying these: $(abc)^2 = 15 	imes 10 	imes 6 = 900 \implies abc = 30$. Then $c = \frac{abc}{ab} = \frac{30}{15} = 2$,$a = \frac{abc}{bc} = \frac{30}{10} = 3$,$b = \frac{abc}{ca} = \frac{30}{6} = 5$. Thus,the expression is $(\sqrt{5} + \sqrt{3} - \sqrt{2})^2$. Taking the square root,we get $\sqrt{5} + \sqrt{3} - \sqrt{2}$.

$\sqrt{10 - \sqrt{24} - \sqrt{40} + \sqrt{60}} = ?$

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