Question

solve this I will mark as brain list












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Answer 1

Answer:

• a = 7
• b = 4

Step-by-step explanation:

We have,

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

Now, rationalising the denominator of LHS

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

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

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

${ \implies \tt \dfrac{ 4 + 3 +4\sqrt{3} }{1} = a + b \sqrt{3} }$

${ \implies \tt7 +4\sqrt{3} = a + b \sqrt{3} }$

On comparing LHS and RHS, we get:

• $\boxed{\sf a = 7 }$
• $\boxed{\sf b = 4}$

Hence the required values of a and b are 7 and 4 respectively.