Question

if A and B are rational number and
$\frac{2 + \sqrt{3} }{2 - \sqrt{3} }$
= a + b √3 find the values of a and b​

Answer 1

$\bf{\large{\underline{\underline{Answer:-}}}}$

$\boxed{ \sf a = 7, \: b = 4}$

$\bf{\large{\underline{\underline{Explanation:-}}}}$

Given :- If a and b are rational numbers $\sf{ \dfrac{2 + \sqrt{3} }{2 - \sqrt{3} } = a + b \sqrt{3} }$

To find :- Values of a and b

Solution :-

$\dfrac{2 + \sqrt{3} }{2 - \sqrt{3} } = a + b \sqrt{3}$

Consider Left Hand Side of the equation

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

To find the values of a and b first we need to get Left Hand Side as Right Hand Side form

So, we shold rationalise the denominator to get Left Hand Side as Right Hand Side form

The rationalising factor of 2 - √3 is 2 + √3 . So multiply both numerator and denominator of the fraction with rationalising factor.

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

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

[Since (x - y)(x + y) = x² - y² Here x = 2, y = √3]

$=\dfrac{ {2}^{2} + 2(2)( \sqrt{3}) + {( \sqrt{3})}^{2}}{4 - 3}$

[Since (x + y)² = x² + 2xy + y² Here x = 2, y = √3]

$= \dfrac{4 + 4 \sqrt{3} + 3 }{1}$

$= \dfrac{7 + 4 \sqrt{3} }{1}$

$= 7 + 4 \sqrt{3}$

Now 7 + 4√4 is in the form of Right Hand Side of the equation ( $\sf{ \dfrac{2 + \sqrt{3} }{2 - \sqrt{3} } = a + b \sqrt{3} }$ )

So, Now we can find values of a and b

$7 + 4 \sqrt{3} = a + b \sqrt{3}$

By equating the corresponding rational and irrational numbers we have,

$a = 7$

$b \sqrt{3} = 4 \sqrt{3}$

$b = 4$

$\Huge{\boxed{ \sf a = 7, \: b = 4}}$

$\bf{\large{\underline{\underline{Identities \: used :-}}}}$

1. (x + y)(x - y) = x² - y²

2. (x + y)² = x² + 2xy + y²

Answer 2

$\large\red{\underline{\boxed{a = 7,b = 4 Answer}}}$

See above attachment.