frac{14}{3+sqrt{2}}=? | vedclass

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

Question

(A) To solve $\frac{14}{3+\sqrt{2}}$,we rationalize the denominator by multiplying the numerator and denominator by the conjugate $(3-\sqrt{2})$.$\fra

Vedclass · Accepted Answer

$3.172$

Vedclass · Answer

(A) To solve $\frac{14}{3+\sqrt{2}}$,we rationalize the denominator by multiplying the numerator and denominator by the conjugate $(3-\sqrt{2})$.$\frac{14}{3+\sqrt{2}} = \frac{14(3-\sqrt{2})}{(3+\sqrt{2})(3-\sqrt{2})}$Using the identity $(a+b)(a-b) = a^2 - b^2$,we get:$= \frac{14(3-\sqrt{2})}{3^2 - (\sqrt{2})^2} = \frac{14(3-\sqrt{2})}{9-2} = \frac{14(3-\sqrt{2})}{7}$$= 2(3-\sqrt{2})$Given $\sqrt{2} \approx 1.414$,we substitute the value:$= 2(3 - 1.414) = 2(1.586) = 3.172$

$\frac{14}{3+\sqrt{2}}=?$

Similar Questions

The square root of $\frac{(0.75)^{3}}{1-0.75} + [0.75 + 0.75^{2} + 1]$ is

If $a=64$ and $b=289,$ then the value of $(\sqrt{\sqrt{a}+\sqrt{b}}-\sqrt{\sqrt{b}-\sqrt{a}})^{\frac{1}{2}}$ is

If $\sqrt{12} = 3.464$,the value of $\sqrt{\frac{3}{4}} + 2 \sqrt{\frac{4}{3}}$ is

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

The expression $(2+\sqrt{2})+\frac{1}{(2+\sqrt{2})}+\frac{1}{(2-\sqrt{2})}$ equals

Vedclass Test Series

Exam Paper Generator

Online Exam Module