The values of a and b if 4+32448−128+200−812+58 is

Given $$4+32448$$−$$128+200$$−$$812+58=4+32122=4+32163$$−$$82+102$$−$$163+102=42+624=4224+624=14+26=a+b2a=14$$,$$b=16$$

Q.

The values of $a and b$ if $\frac{4 + 3 \sqrt{2}}{4 \sqrt{48} - \sqrt{128} + \sqrt{200} - 8 \sqrt{12} + 5 \sqrt{8}}$ is

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By Expert Faculty of Sri Chaitanya

$a = \frac{1}{4}, b = \frac{1}{6}$

$a = \frac{1}{4}, b = - \frac{1}{6}$

$a = - \frac{1}{4}, b = - \frac{1}{6}$

$a = - \frac{1}{4}, b = \frac{1}{6}$

answer is A.

(Detailed Solution Below)

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Detailed Solution

Given $\frac{4 + 3 \sqrt{2}}{4 \sqrt{48} - \sqrt{128} + \sqrt{200} - 8 \sqrt{12} + 5 \sqrt{8}}$

$= \frac{4 + 3 \sqrt{2}}{12 \sqrt{2}}$

$= \frac{4 + 3 \sqrt{2}}{16 \sqrt{3} - 8 \sqrt{2} + 10 \sqrt{2} - 16 \sqrt{3} + 10 \sqrt{2}}$

$= \frac{4 \sqrt{2} + 6}{24}$

$= \frac{4 \sqrt{2}}{24} + \frac{6}{24}$

$= \frac{1}{4} + \frac{\sqrt{2}}{6}$

$= a + b \sqrt{2}$

$a = \frac{1}{4}, b = \frac{1}{6}$

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