sqrt {[10 - sqrt {24} - sqrt {40} + sqrt {60} | vedclass

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(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\sq

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: $a + b + c + 2\sqrt {ab} - 2\sqrt {bc} - 2\sqrt {ac}$. Comparing with $10 - 2\sqrt {6} - 2\sqrt {10} + 2\sqrt {15}$,we have $a+b+c = 10$. Also,$2\sqrt {ab} = 2\sqrt {15} \implies ab = 15$,$2\sqrt {bc} = 2\sqrt {10} \implies bc = 10$,$2\sqrt {ac} = 2\sqrt {6} \implies ac = 6$. Multiplying these: $(abc)^2 = 15 	imes 10 	imes 6 = 900$,so $abc = 30$. Then $c = \frac{abc}{ab} = \frac{30}{15} = 2$,$a = \frac{abc}{bc} = \frac{30}{10} = 3$,$b = \frac{abc}{ac} = \frac{30}{6} = 5$. Thus,the expression is $(\sqrt {3} + \sqrt {5} - \sqrt {2})^2$. Therefore,$\sqrt {10 - \sqrt {24} - \sqrt {40} + \sqrt {60}} = \sqrt {5} + \sqrt {3} - \sqrt {2}$.

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

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