Question

If √4 + √3/2√4 -√3 = a - b √12 then find a and b​

Answer 1

Answer:

234-6568

&+753589

Step-by-step explanation:

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Answer 2

$\sf\large\underline\blue{Given:-}$

$\\ \: \: { \huge{.}} \: \: { \bold{ \dfrac{ \sqrt{4} + \sqrt{3} }{2 \sqrt{4} - \sqrt{3} } = a - b \sqrt{12} }}$

$\sf\large\underline\blue{To\: Find:-}$

• Value of a and b

$\sf\large\underline\blue{Formula\: used:-}$

$\\ \: \: \twoheadrightarrow \: { \bold{ (a + b)(a - b) = {a}^{2} - {b}^{2} }}$

$\sf\large\underline\blue{Solution:-}$

Given that,

$↦{ \bold{ \dfrac{ \sqrt{4} + \sqrt{3} }{2 \sqrt{4} - \sqrt{3} } = a - b \sqrt{12} }}$

( Rationalization of denominator )

$↦{ \bold{ \dfrac{ \sqrt{4} + \sqrt{3} }{2 \sqrt{4} - \sqrt{3} } \times \dfrac{2 \sqrt{4} + \sqrt{3} }{2 \sqrt{4} + \sqrt{3} } = a - b \sqrt{12} }}$

$↦ { \bold{ \dfrac{( \sqrt{4} + \sqrt{3})(2 \sqrt{4} + \sqrt{3})}{(2 \sqrt{4} - \sqrt{3})(2 \sqrt{4} + \sqrt{3})} = a - b \sqrt{12} }}$

$↦{ \bold{ \dfrac{( \sqrt{4} + \sqrt{3})(2 \sqrt{4} + \sqrt{3})}{(2 \sqrt{4})^{2} - ( \sqrt{3})^{2} } = a - b \sqrt{12} }}$

$↦{ \bold{ \dfrac{ \{2 {( \sqrt{4} )}^{2} + \sqrt{4} \sqrt{3} + 2\sqrt{3} \sqrt{4} + {( \sqrt{3}) }^{2} \}}{(2 \sqrt{4})^{2} - ( \sqrt{3})^{2} } = a - b \sqrt{12} }}$

$↦{ \bold{ \dfrac{ \{8 + \sqrt{12}+ 2\sqrt{12}+3 \}}{16 - 3} = a - b \sqrt{12} }}$

$↦ { \bold{ \dfrac{11 + 3\sqrt{12}}{13} = a - b \sqrt{12} }}$

$↦ { \bold{ \dfrac{11}{13} + \dfrac{3\sqrt{12}}{13} = a - b \sqrt{12} }}$

After comparing, we get

$\boxed{ ↦a = \frac{11}{13}} \\ \\ \boxed{ ↦b = - \frac{3}{13}}</p><p>$