Question

√4+√3/2√4-√3=a-b√12 ​

Answer 1

GIVEN :–

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

TO FIND :–

• a and b = ?

SOLUTION :–

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

• Rationalization of denominator –

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

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

• Using identity –

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

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

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

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

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

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

▪︎ Now compare –

$\\ \: \: \dashrightarrow \: \: \large { \boxed{ \bold{a = \dfrac{11}{13}}}}$

• And –

$\\ \: \: \dashrightarrow \: \: \large { \boxed{ \bold{b= - \dfrac{3}{13}}}}$

Answer 2

Step-by-step explanation:

Step-by-step explanation:

${\red{\underline{\underline{\bold{Given:-}}}}}$

$\frac{4 + \sqrt{3} }{2\sqrt{4} - \sqrt{3} } = a - b \sqrt{12}$

${\blue{\underline{\underline{\bold{To\:Find:-}}}}}$

• The values of ' a ' and ' b '

$\tt\orange{Concept \: used\:here:-}$

• Rationalising of the numbers.

${\green{\underline{\underline{\bold{Solution:-}}}}}$

$\frac{4 + \sqrt{3} }{4- \sqrt{3} } = a - b \sqrt{12}$

By rationalising:-

$\frac{4 + \sqrt{3} }{2 \sqrt{4} - \sqrt{3} } \times \frac{2 \sqrt{4} + \sqrt{3} }{2 \sqrt{4} + \sqrt{3} } = a - b \sqrt{12} \\ \\ \implies \frac{ {(4 + \sqrt{3} )}(2 \sqrt{4} + 3) }{ ({2 \sqrt{4} )}^{2} - {( \sqrt{3} )}^{2} } = a - b \sqrt{12} \: \: \: \: [(a+b) (a-b) = {a}^{2}-{b}^{2}]\\ \\ \implies \frac{ 2{ (\sqrt{4}) }^{2} + \sqrt{4} \sqrt{3} + 2( \sqrt{4} )( \sqrt{3}) + ({ \sqrt{3}) }^{3} }{16-3} = a - b \sqrt{12}\: \: \: \: \\ \\ \implies \frac{8 + \sqrt{12} + 2\sqrt{12} + 3 }{ 13} = a - b \sqrt{12} \\ \\ \implies \frac{11 + 3 \sqrt{12} }{13} = a - b \sqrt{12} \\ \\ \implies \frac{11}{13} + \frac{3 \sqrt{12} }{13} = a - b \sqrt{12}$

Now comparing the values of $\frac{11}{13} + \frac{3 \sqrt{12} }{13}$ with $a - b\sqrt{12}$

$\boxed{\implies a = \frac{11}{13}}$

$\boxed{\implies b = - \frac{3}{13}}$