Question

if 5-2root 3÷7-4root3=a-root 3 b,then find a and b​

Answer 1

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

$\Large{a = 11, \: b = - 6}$

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

Given :- $\sf{ \dfrac{5 - 2 \sqrt{3} }{7 - 4 \sqrt{3} } = a - \sqrt{3}b}$

To find :- Value of a and b

Solution :-

$\dfrac{5 - 2 \sqrt{3} }{7 - 4 \sqrt{3} } = a - \sqrt{3}b$

Consider Left Hand Side

$\dfrac{5 - 2 \sqrt{3} }{7 - 4 \sqrt{3}}$

Now to rationalise the denominator

The rationalising factor of 7 - 4√3 is 7 + 4√3. So, multiply both numerator and denominator by Rationalising factor.

$= \dfrac{5 - 2 \sqrt{3} }{7 - 4 \sqrt{3} } \times \dfrac{7 + 4 \sqrt{3} }{7 + 4 \sqrt{3} }$

$= \dfrac{5(7 + 4 \sqrt{3}) - 2 \sqrt{3}(7 + 4 \sqrt{3}) }{ {7}^{2} - {(4 \sqrt{3})}^{2} }$

[Since (x - y)(x + y) = x² - y² and here x = 7, y = 4√3]

$= \dfrac{35 + 20 \sqrt{3} - 14 \sqrt{3} - 8(3) }{49 - {4}^{2} {(\sqrt{3}) }^{2} }$

[Here √3 * √3 = √9 = 3]

$= \dfrac{35 +6 \sqrt{3} - 24}{49 - 16(3)}$

$= \dfrac{11 + 6\sqrt{3} }{1}$

$= 11 + 3 \sqrt{3}$

Now consider, $\sf{ \dfrac{5 - 2 \sqrt{3} }{7 - 4 \sqrt{3} } = a - \sqrt{3}b}$

Equating to the corresponding rational and irrational we have

a = 11

- √3 b = 6√3

- b = 6

b = - 6

$\boxed{\bf{a = 11, \: b = -6}}$

$\bf{\underline{\underline{Note:-}}}$

(x + y)(x - y) = x² - y² or (x - y)(x + y) = x² - y² are same.