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Colt1911 [192]
3 years ago
9

If √3 = 1.732, then √[(√3-1)/(√3+1)] is equal to​

Mathematics
1 answer:
adelina 88 [10]3 years ago
3 0

Step-by-step explanation:

<h3><u>Given Question :-</u></h3>

\sf \:  \sqrt{3} = 1.732, \: then \:  \sqrt{\dfrac{ \sqrt{3}  - 1}{ \sqrt{3}  + 1} }

\red{\large\underline{\sf{Solution-}}}

Given expression is

\rm :\longmapsto\: \sqrt{\dfrac{ \sqrt{3}  - 1}{ \sqrt{3}  + 1} }

<em>On rationalizing the denominator, we get </em>

\rm \:  =  \:  \sqrt{\dfrac{ \sqrt{3}  - 1}{ \sqrt{3}  + 1}  \times \dfrac{ \sqrt{3}  - 1}{ \sqrt{3}  - 1} }

\rm \:  =  \:  \sqrt{\dfrac{ {( \sqrt{3}  - 1)}^{2} }{( \sqrt{3}  + 1)( \sqrt{3}  - 1)} }

<em>We know, </em>

\boxed{ \tt \: (x - y)(x + y) =  {x}^{2} -  {y}^{2} \: }

<em>So, using this, we get </em>

\rm \:  =  \: \dfrac{ \sqrt{3}  - 1}{ \sqrt{ {( \sqrt{3})}^{2}  -  {(1)}^{2} } }

\rm \:  =  \: \dfrac{ \sqrt{3}  - 1}{ \sqrt{ 3 - 1} }

\rm \:  =  \: \dfrac{ \sqrt{3}  - 1}{ \sqrt{2} }

\rm \:  =  \: \dfrac{ \sqrt{3}  - 1}{ \sqrt{2} } \times \dfrac{ \sqrt{2} }{ \sqrt{2} }

\rm \:  =  \: \dfrac{(1.732 - 1) \sqrt{2} }{2}

\rm \:  =  \: \dfrac{0.732 \times \sqrt{2} }{2}

\rm \:  =  \: 0.366 \sqrt{2}

Hence,

\rm :\longmapsto\: \boxed{ \rm{ \: \:  \:   \sqrt{\dfrac{ \sqrt{3}  - 1}{ \sqrt{3}  + 1} } = 0.366 \sqrt{2} \:  \:  \: }}

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

<h2>More Identities to know :- </h2>

(a + b)² = a² + 2ab + b²

(a - b)² = a² - 2ab + b²

a² - b² = (a + b)(a - b)

(a + b)² = (a - b)² + 4ab

(a - b)² = (a + b)² - 4ab

(a + b)² + (a - b)² = 2(a² + b²)

(a + b)³ = a³ + b³ + 3ab(a + b)

(a - b)³ = a³ - b³ - 3ab(a - b)

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