Question

rationalize the denominator:- 1/ root3 - root 2 + 1

Answer 1

GIVEN :

Rationalize the denominator of  $\frac{1}{\sqrt{3}-\sqrt{2} + 1}$

TO RATIONALIZE :

The denominator for the given expression .

SOLUTION :

Given expression is $\frac{1}{\sqrt{3}-\sqrt{2} + 1}$

Now solving the $\frac{1}{\sqrt{3}-\sqrt{2} + 1}$

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

$=\frac{1}{\sqrt{3}-(\sqrt{2} + 1)}$

Now to rationalize the denominator by multiplying and dividing by the denominator's conjugate,

$=\frac{1}{\sqrt{3}-(\sqrt{2} + 1)}\times \frac{\sqrt{3}+(\sqrt{2} + 1)}{\sqrt{3}+(\sqrt{2} + 1)}$

$=\frac{1(\sqrt{3})+(\sqrt{2} + 1)}{(\sqrt{3})^2-(\sqrt{2}+1)^2}$

By using the Algebraic identities :

$(a+b)^2=a^2+b^2+2ab$

$(a+b)(a-b)=a^2-b^2$

$=\frac{\sqrt{3}+(\sqrt{2} + 1)}{3-(\sqrt{2})^2+(1)^2+2(\sqrt{2})(1)}$

$=\frac{\sqrt{3}+(\sqrt{2} + 1)}{3-2-1-2\sqrt{2}}$

$=\frac{\sqrt{3}+(\sqrt{2} + 1)}{-2\sqrt{2}}$

Now multiply and divide by $\sqrt{2}$ we get,

$=\frac{\sqrt{3}+(\sqrt{2} + 1)}{-2\sqrt{2}}\times \frac{\sqrt{2}}{\sqrt{2}}$

$=\frac{\sqrt{3}+(\sqrt{2} + 1)(\sqrt{2})}{-2(2)}$

$=\frac{\sqrt{3}(\sqrt{2})+(\sqrt{2})(\sqrt{2}) + (1)(\sqrt{2})}{-4}$

$=\frac{\sqrt{6}+2 +\sqrt{2}}{-4}$

$=-(\frac{\sqrt{6}+2 +\sqrt{2}}{4})$

∴ the given expression is rationalized into the expression is $-(\frac{\sqrt{6}+2 +\sqrt{2}}{4})$

∴ $\frac{1}{\sqrt{3}-\sqrt{2} + 1}=-(\frac{\sqrt{6}+2 +\sqrt{2}}{4})$

Answer 2

Step-by-step explanation:

-(√6+2√2)/4 answer of this question