Question

(sqrt3-1)/(sqrt3+1)=a+bsqrt3

Answer 1

Solution :

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Given :

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

a & b both are rational numbers.
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To Find :

a possible values for a & b

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We can find a possible value by,

⇒ Rationalizing the denominator,.

By using the principle (only for integers in square root) :

(a - b) (a + b) = a² - b² ,  => in denominator (To bring them to standard form)

Here the denominator :

⇒ √3 + 1 is in the form a + b

So, Multiplying both numerator & denominator by √3 - 1,. as (a - b)

We will get the form (a + b)(a - b)

⇒ By this way we can find values easily,.

The steps are as Follows :

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

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

The numerator is of the form :

(a - b)²,.

Hence,.

We know that,.

(a - b)² = a² - 2ab + b²

By substituting the values ,

We get,.

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

⇒ $\frac{3 + 1 -2 \sqrt{3} }{3 - 1}$

⇒ $\frac{4 - 2 \sqrt{3} }{2}$

⇒ $\frac{2(2 - \sqrt{3} )}{2}$

⇒ 2 - √3

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It is of the form a + b√3

Hence,.

if a = 2 & b = -1,.

⇒ 2 + (-1)√3

⇒ 2 - √3

∴ a = 2,

∴ b = -1

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                                       Hope it Helps !!

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Answer 2

Hey friend,
Here is the answer you were looking for:
$\frac{ \sqrt{3} - 1}{ \sqrt{3} + 1 } = a + b \sqrt{3} \\ \\ on \: rationalizing \: the \: denominator \: we \: get \\ \\ = \frac{ \sqrt{3} - 1 }{ \sqrt{3} + 1 } \times \frac{ \sqrt{3} - 1 }{ \sqrt{3} - 1 } \\ \\ using \: the \: identities \\ {(a - b)}^{2} = {a}^{2} + {b}^{2} - 2ab \\ (a + b)(a - b) = {a}^{2} - {b}^{2} \\ \\ = \frac{ {( \sqrt{3}) }^{2} + {(1)}^{2} - 2 \times \sqrt{3} \times 1}{ {( \sqrt{3}) }^{2} - {(1)}^{2} } \\ \\ = \frac{3 + 1 - 2 \sqrt{3} }{3 - 1} \\ \\ = \frac{4 - 2 \sqrt{3} }{2} \\ \\ 2 - \sqrt{3} = a + b \sqrt{3} \\ \\ a = 2 \\ \\ b = - 1$


Hope this helps!!!

@Mahak24

Thanks...
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