Question

Find the values of a and b.

Answer 1

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Answer 2

Hey friend, Harish here.

Here is your answer:

Given that,

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

To find,

The value of a & b.

Solution,

First we must rationalize the denominator of the number by multiplying and dividing it by it's conjugate.

The conjugate of the denominator is √3 - 1.

Then, 

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

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

We know that, (a+b)(a-b)= a² - b².

So, (√3+1)(√3-1) = (√3)² - 1² = 3-1 = 2.

Then,

$\frac{ (\sqrt{3}-1)^{2} }{ 2} = \frac{ 3+1-2( \sqrt{3})(1) }{ 2} = \frac{4-2 \sqrt{2} }{2} = 2 - \sqrt{3} =a+b \sqrt{3}$

Now, By comparing we get, a = 2 & b = -1.
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Hope my answer is helpful to you.