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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Similar Questions

If ${\left( \frac{2}{3} \right)^{x + 2}} = {\left( \frac{3}{2} \right)^{2 - 2x}}$,then $x =$

If $12^{4+2x^2} = (24\sqrt{3})^{3x^2-2}$,then $x$ is equal to

Which of the following statements are true and which are false? In each case,give a valid reason for your answer.
$t: \sqrt{11}$ is a rational number.

If ${x^{x \cdot \sqrt[3]{x}}} = {(x \cdot \sqrt[3]{x})^x}$,then $x =$

$\frac{3\sqrt{2}}{\sqrt{6} + \sqrt{3}} - \frac{4\sqrt{3}}{\sqrt{6} + \sqrt{2}} + \frac{\sqrt{6}}{\sqrt{3} + \sqrt{2}} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module