Question

If 5+2 root 3/7+root 3= a-root 3b, find a and b where a and b are rational numbers.

Answer 1

Answer:

$\large{\underline{\boxed{\sf a = \frac{29}{46} \: and \: b = - \frac{9}{46}}}}$

Step-by-step explanation:

$\sf \implies \frac{5 + 2 \sqrt{3} }{7 + \sqrt{3} } = a - \sqrt{3} b \\ \\ \sf \implies \frac{5 + 2 \sqrt{3} }{7 + \sqrt{3} } \times \frac{7 - \sqrt{3} }{7 - \sqrt{3} } \\ \\ \sf \implies \frac{(5 + 2 \sqrt{3})(7 - \sqrt{3})}{(7)^{2} - { (\sqrt{3} )}^{2} } \\ \\ \sf \implies \frac{35 - 5 \sqrt{3} + 14 \sqrt{3} - 6 }{49 - 3} \\ \\ \sf \implies \frac{29 + 9 \sqrt{3} }{46} \\ \\ \sf \implies \frac{29}{46} + \frac{9 }{46} \sqrt{3}$

On comparing it with a - √3 b,

$\boxed{ \bf a = \frac{29}{46} \: and \: b = - \frac{9}{46} }$

_______________________

$\large{\underline{\underline{\mathfrak{\heartsuit \: Extra \: Information: \: \heartsuit}}}}$

To rationalise the denominator, just multiply the whole fraction by its rationalising factor.

Example -

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

Here, the rationalising factor of 2 - √3 is 2 + √3.

Answer 2

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

$\boxed{\bold{a = \dfrac{29}{46}, \: b = - \dfrac{9}{46}}}$

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

Given :

$\dfrac{5 + 2\sqrt{3} }{7 + \sqrt{3} } = a - \sqrt{3}b$

To find : Value of a and b

Solution :

$\sf{\dfrac{5 + 2\sqrt{3} }{7 + \sqrt{3}} = a - \sqrt{3}b}$

Consider Left Hand Side

$\sf{\dfrac{5 + 2\sqrt{3} }{7 + \sqrt{3}}}$

Rationalize the denominator

The Rationalizing factor of 7 + √3 is 7 - √3 . So, Multiply numerator and denominator with rationalising factor.

$= \dfrac{5 + 2 \sqrt{3} }{7 + \sqrt{3} } \times \dfrac{7 - \sqrt{3} }{7 - \sqrt{3} }$

$= \dfrac{5(7 - \sqrt{3}) + 2 \sqrt{3}(7 - \sqrt{3})}{ {7}^{2} - {( \sqrt{3})}^{2} }$

Since (x + y)(x - y) = x² - y²

$= \dfrac{35 - 5 \sqrt{3} +14 \sqrt{3} - 2(3)}{49 - 3}$

$= \dfrac{35 + 9 \sqrt{3} - 6 }{46}$

$= \dfrac{29 + 9 \sqrt{3} }{46}$

$= \dfrac{29}{46} + \dfrac{9 \sqrt{3} }{46}$

Now Consider, $\dfrac{5 + 2\sqrt{3} }{7 + \sqrt{3} } = a - \sqrt{3}b$

$\sf{\implies{ \dfrac{29}{46} + \dfrac{9 \sqrt{3} }{46}= a - \sqrt{3}b}}$

Equating to the corresponding rational and irrational number we have ,

$a = \dfrac{29}{46}$

$- \sqrt{3}b = \dfrac{9 \sqrt{3} }{46}$

$- b = \dfrac{9}{46}$

$b = - \dfrac{9}{46}$

$\boxed{\bold{a = \dfrac{29}{46}, \: b = - \dfrac{9}{46}}}$

Identity used :-

(x + y)(x - y) = x² - y²

Extra info related to the question :-

What is a rationlising factor ?

If the product of two irrational numbers is a rational number then each of the the two is rationalising factor of the other.

How to rationalise the denominator ?

1) Consider the denominator of the fraction.

2) Then multiply the numerator and denominator with rationalising factor of the denominator.