Given sqrt{5}=2.2361 and sqrt{3}=1.7321,then | vedclass

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

Question

(C) To solve $\frac{1}{\sqrt{5}-\sqrt{3}}$,we rationalize the denominator by multiplying the numerator and the denominator by the conjugate $(\sqrt{5}

Vedclass · Accepted Answer

$1.9841$

Vedclass · Answer

(C) To solve $\frac{1}{\sqrt{5}-\sqrt{3}}$,we rationalize the denominator by multiplying the numerator and the denominator by the conjugate $(\sqrt{5}+\sqrt{3})$.$\frac{1}{\sqrt{5}-\sqrt{3}} = \frac{1}{(\sqrt{5}-\sqrt{3})} 	imes \frac{(\sqrt{5}+\sqrt{3})}{(\sqrt{5}+\sqrt{3})}$Using the identity $(a-b)(a+b) = a^2 - b^2$ in the denominator:$= \frac{\sqrt{5}+\sqrt{3}}{(\sqrt{5})^2 - (\sqrt{3})^2}$$= \frac{\sqrt{5}+\sqrt{3}}{5-3} = \frac{\sqrt{5}+\sqrt{3}}{2}$Substitute the given values $\sqrt{5} = 2.2361$ and $\sqrt{3} = 1.7321$:$= \frac{2.2361 + 1.7321}{2} = \frac{3.9682}{2} = 1.9841$

Given $\sqrt{5}=2.2361$ and $\sqrt{3}=1.7321$,then $\frac{1}{\sqrt{5}-\sqrt{3}} = $

Similar Questions

$(15.01)^{2} \times \sqrt{730} = ?$

The smallest number which is a perfect square and contains $7936$ as a factor is

Find the square root of $211600$.

If $\frac{4 \sqrt{3}+5 \sqrt{2}}{\sqrt{48}+\sqrt{18}}=a+b \sqrt{6},$ then the values of $a$ and $b$ are,respectively

In the following questions,two equations numbered $I$ and $II$ are given. You have to solve both the equations and give the answer.
$I$: $\sqrt{25 x^{2}} - 125 = 0$
$II$: $\sqrt{361} y + 95 = 0$

Vedclass Test Series

Exam Paper Generator

Online Exam Module