If x=frac{sqrt{3}+sqrt{2}}{sqrt{3}-sqrt{2}} a | vedclass

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

Question

(C) Rationalizing the denominator for $x$:$x = \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}} 	imes \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}} = \frac{

Vedclass · Accepted Answer

$98$

Vedclass · Answer

(C) Rationalizing the denominator for $x$:$x = \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}} 	imes \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}} = \frac{(\sqrt{3}+\sqrt{2})^{2}}{3-2} = 3+2+2\sqrt{6} = 5+2\sqrt{6}$.Similarly,for $y$:$y = \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}} 	imes \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}} = \frac{(\sqrt{3}-\sqrt{2})^{2}}{3-2} = 3+2-2\sqrt{6} = 5-2\sqrt{6}$.Now,calculate $x+y$ and $xy$:$x+y = (5+2\sqrt{6}) + (5-2\sqrt{6}) = 10$.$xy = (5+2\sqrt{6})(5-2\sqrt{6}) = 5^{2} - (2\sqrt{6})^{2} = 25 - 24 = 1$.Using the identity $x^{2}+y^{2} = (x+y)^{2} - 2xy$:$x^{2}+y^{2} = (10)^{2} - 2(1) = 100 - 2 = 98$.

If $x=\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}$ and $y=\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}},$ then find the value of $x^{2}+y^{2}$.

Similar Questions

Rationalise the denominator in each of the following and hence evaluate by taking $\sqrt{2}=1.414, \sqrt{3}=1.732$ and $\sqrt{5}=2.236,$ up to three decimal places.
$\frac{\sqrt{10}-\sqrt{5}}{2}$

The decimal expansion of the number $\sqrt{2}$ is

Insert a rational number and an irrational number between the following:
$0.0001$ and $0.001$

State whether the following statement is true or false. Justify your answer.
$\frac{\sqrt{15}}{\sqrt{3}}$ is written in the form $\frac{p}{q},$ where $q \neq 0,$ so it is a rational number.

Simplify $: \frac{(25)^{\frac{3}{2}} \times (243)^{\frac{3}{5}}}{(16)^{\frac{5}{4}} \times (8)^{\frac{4}{3}}}$

Vedclass Test Series

Exam Paper Generator

Online Exam Module