sqrt{frac{4}{3}}-sqrt{frac{3}{4}}=? | vedclass

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

Question

(A) To solve the expression $\sqrt{\frac{4}{3}}-\sqrt{\frac{3}{4}}$,we first simplify the square roots of the individual terms.$\sqrt{\frac{4}{3}} = \

Vedclass · Accepted Answer

$\frac{1}{2 \sqrt{3}}$

Vedclass · Answer

(A) To solve the expression $\sqrt{\frac{4}{3}}-\sqrt{\frac{3}{4}}$,we first simplify the square roots of the individual terms.$\sqrt{\frac{4}{3}} = \frac{\sqrt{4}}{\sqrt{3}} = \frac{2}{\sqrt{3}}$$\sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{\sqrt{4}} = \frac{\sqrt{3}}{2}$Now,subtract the two fractions:$\frac{2}{\sqrt{3}} - \frac{\sqrt{3}}{2} = \frac{2 	imes 2 - \sqrt{3} 	imes \sqrt{3}}{2 \sqrt{3}}$$= \frac{4 - 3}{2 \sqrt{3}}$$= \frac{1}{2 \sqrt{3}}$

$\sqrt{\frac{4}{3}}-\sqrt{\frac{3}{4}}=?$

Similar Questions

The value of $\sqrt{\frac{(0.03)^{2}+(0.21)^{2}+(0.065)^{2}}{(0.003)^{2}+(0.021)^{2}+(0.0065)^{2}}}$ is

What should come in place of both the question marks in the equation $\frac{?}{\sqrt{128}} = \frac{\sqrt{162}}{?}$?

$\sqrt{1.5625} = ?$

$\frac{1}{\sqrt{9}-\sqrt{8}} = ?$

The digit in the unit's place in the square root of $15876$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module