Question

Rationalize the denominator.
$\frac{1}{2 + \sqrt{ 3} }$
​

Answer 1

Answer:

we will multiply both the numerator and denominator by 2-√3

so we will get 2-√3

Answer 2

Answer:

$\dfrac{1}{2+\sqrt{3}}=2-\sqrt{3}$

Step-by-step explanation:

$\dfrac{1}{2+\sqrt{3}}$

$\mathrm{Multiply\:by\:the\:conjugate}\:\dfrac{2-\sqrt{3}}{2-\sqrt{3}}$

$=\dfrac{1\cdot \left(2-\sqrt{3}\right)}{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}$

$1\cdot \left(2-\sqrt{3}\right)=2-\sqrt{3}$

$\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)$

$\mathrm{Apply\:Difference\:of\:Two\:Squares\:Formula:\:}\left(a+b\right)\left(a-b\right)=a^2-b^2$ $a=2,\:b=\sqrt{3}$

$=2^2-\left(\sqrt{3}\right)^2$

$\mathrm{Simplify}\:2^2-\left(\sqrt{3}\right)^2$

$=4-3$

$\mathrm{Subtract\:the\:numbers:}\:4-3=1$

$=1$

$=\dfrac{2-\sqrt{3}}{1}$

$\mathrm{Apply\:rule}\:\dfrac{a}{1}=a$

$=2-\sqrt{3}$