Question

Find the value of the following (explain too)
$\frac{\sqrt{3} +1}{\sqrt{3} -1} = a+b\sqrt{3}$$\frac{\sqrt{3} +1}{\sqrt{3} -1} = a+b\sqrt{3}$

Answer 1

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We Have,

$= \frac{ \sqrt{3} + 1 }{ \sqrt{3} - 1 } - \frac{ \sqrt{3} - 1}{ \sqrt{3} - 1 } = a + b \sqrt{3}$

$⇒ \frac{( \sqrt{3} + 1 {)}^{2} - ( \sqrt{3} - 1 {)}^{2} }{( \sqrt{3} - 1)( \sqrt{3} + 1) } = a + b \sqrt{3}$

$\[ \left[ \text{Using Formula = } {a}^{2} - {b}^{2} = (a + b)(a - b)\right] \]$

$⇒ \frac{( \sqrt{3} + 1 + \sqrt{3} - 1)( \sqrt{3} + 1 - \sqrt{3} + 1) }{(( \sqrt{3} {)}^{2} - {1}^{2} ) }$

$⇒ \frac{2 \sqrt{3} \times 2 }{3 - 1} = a + b \sqrt{3}$

$⇒2 \sqrt{3} = a + b \sqrt{3}$

$\textbf{Comparing L.H.S Ans R.H.S we get,}$

$b = 2 \: \textbf{and} \: a = 0$

Thus Value of a and b are 0 and 2 respectively.

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Answer 2

Step-by-step explanation:

=

3

−1

3

+1

−

3

−1

3

−1

=a+b

3

⇒ \frac{( \sqrt{3} + 1 {)}^{2} - ( \sqrt{3} - 1 {)}^{2} }{( \sqrt{3} - 1)( \sqrt{3} + 1) } = a + b \sqrt{3}⇒

(

3

−1)(

3

+1)

(

3

+1)

2

−(

3

−1)

2

=a+b

3

\left[ \text{Using Formula = } {a}^{2} - {b}^{2} = (a + b)(a - b)\right][Using Formula = a

2

−b

2

=(a+b)(a−b)]

⇒ \frac{( \sqrt{3} + 1 + \sqrt{3} - 1)( \sqrt{3} + 1 - \sqrt{3} + 1) }{(( \sqrt{3} {)}^{2} - {1}^{2} ) }⇒

((

3

)

2

−1

2

)

(

3

+1+

3

−1)(

3

+1−

3

+1)

⇒ \frac{2 \sqrt{3} \times 2 }{3 - 1} = a + b \sqrt{3}⇒

3−1

2

3

×2

=a+b

3

⇒2 \sqrt{3} = a + b \sqrt{3}⇒2

3

=a+b

3

\textbf{Comparing L.H.S Ans R.H.S we get,}Comparing L.H.S Ans R.H.S we get,

b = 2 \: \textbf{and} \: a = 0b=2anda=0