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

Sagot :

SalmonWorldTopper

Step-by-step explanation:

Given Question :-

[tex] \sf \: \sqrt{3} = 1.732, \: then \: \sqrt{\dfrac{ \sqrt{3} - 1}{ \sqrt{3} + 1} } [/tex]

[tex] \red{\large\underline{\sf{Solution-}}}[/tex]

Given expression is

[tex]\rm :\longmapsto\: \sqrt{\dfrac{ \sqrt{3} - 1}{ \sqrt{3} + 1} } [/tex]

On rationalizing the denominator, we get

[tex]\rm \:  =  \: \sqrt{\dfrac{ \sqrt{3} - 1}{ \sqrt{3} + 1} \times \dfrac{ \sqrt{3} - 1}{ \sqrt{3} - 1} } [/tex]

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

We know,

[tex] \boxed{ \tt \: (x - y)(x + y) = {x}^{2} - {y}^{2} \: }[/tex]

So, using this, we get

[tex]\rm \:  =  \: \dfrac{ \sqrt{3} - 1}{ \sqrt{ {( \sqrt{3})}^{2} - {(1)}^{2} } } [/tex]

[tex]\rm \:  =  \: \dfrac{ \sqrt{3} - 1}{ \sqrt{ 3 - 1} } [/tex]

[tex]\rm \:  =  \: \dfrac{ \sqrt{3} - 1}{ \sqrt{2} } [/tex]

[tex]\rm \:  =  \: \dfrac{ \sqrt{3} - 1}{ \sqrt{2} } \times \dfrac{ \sqrt{2} }{ \sqrt{2} } [/tex]

[tex]\rm \:  =  \: \dfrac{(1.732 - 1) \sqrt{2} }{2} [/tex]

[tex]\rm \:  =  \: \dfrac{0.732 \times \sqrt{2} }{2} [/tex]

[tex]\rm \:  =  \: 0.366 \sqrt{2} [/tex]

Hence,

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

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More Identities to know :-

(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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