The rationalising factor of 2sqrt{3} - sqrt{7 | vedclass

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(B) To find the rationalising factor of $2\sqrt{3} - \sqrt{7}$,we use the identity $(a - b)(a + b) = a^2 - b^2$.Let $a = 2\sqrt{3}$ and $b = \sqrt{7}$

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) To find the rationalising factor of $2\sqrt{3} - \sqrt{7}$,we use the identity $(a - b)(a + b) = a^2 - b^2$.Let $a = 2\sqrt{3}$ and $b = \sqrt{7}$.The expression is of the form $a - b$.Multiplying by $(a + b) = 2\sqrt{3} + \sqrt{7}$ gives:$(2\sqrt{3} - \sqrt{7})(2\sqrt{3} + \sqrt{7}) = (2\sqrt{3})^2 - (\sqrt{7})^2$$= (4 	imes 3) - 7 = 12 - 7 = 5$.Since $5$ is a rational number,the rationalising factor is $2\sqrt{3} + \sqrt{7}$.

The rationalising factor of $2\sqrt{3} - \sqrt{7}$ is

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