Find the square root of 3-sqrt{5}. | vedclass

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

Question

(A) To find the square root of $3-\sqrt{5}$,we write it as $\sqrt{3-\sqrt{5}}$.Multiply and divide by $\sqrt{2}$ inside the square root:$\sqrt{\frac{2

Vedclass · Accepted Answer

$\sqrt{\frac{5}{2}}-\frac{1}{\sqrt{2}}$

Vedclass · Answer

(A) To find the square root of $3-\sqrt{5}$,we write it as $\sqrt{3-\sqrt{5}}$.Multiply and divide by $\sqrt{2}$ inside the square root:$\sqrt{\frac{2(3-\sqrt{5})}{2}} = \sqrt{\frac{6-2\sqrt{5}}{2}}$.Now,express the numerator $6-2\sqrt{5}$ as a perfect square:$6-2\sqrt{5} = 5 + 1 - 2\sqrt{5} = (\sqrt{5})^2 + (1)^2 - 2(\sqrt{5})(1) = (\sqrt{5}-1)^2$.Substituting this back,we get:$\sqrt{\frac{(\sqrt{5}-1)^2}{2}} = \frac{\sqrt{5}-1}{\sqrt{2}} = \frac{\sqrt{5}}{\sqrt{2}} - \frac{1}{\sqrt{2}} = \sqrt{\frac{5}{2}} - \frac{1}{\sqrt{2}}$.

Find the square root of $3-\sqrt{5}$.

Similar Questions

The decimal expansion of the rational number $\frac{14587}{1250}$ will terminate after

Find the square root of $24+2 \sqrt{119}$.

Prove that $\frac{1}{\sqrt{2}}$ is an irrational number.

Show that the square of any odd integer is of the form $4q + 1$ for some integer $q$.

Prove that the following number is irrational: $2+\sqrt{3}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module