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Algebra of Derivatives

If √ 13+4 √3=...

Question

If $\sqrt$ (13+4 $\sqrt$ 3)= $\sqrt$ 3a+b, then the value of a-b is ?

A

2

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B

1

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C

4

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D

3

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E

None of these

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Solution

The correct option is B
1

${\sqrt (13 + 4 \sqrt 3) = (2 \sqrt{3})}^{2} + 1 + 2 \times (2 \sqrt{3}) \times 1$

$2 \sqrt{3} + 1 = a \sqrt{3} + b$

$⟹ a - b = 2 - 1 = 1$

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