Math, asked by iglegitmax, 1 month ago

if x=7+4√3, find √x+(1/√x)​

Answers

Answered by mathdude500
8

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

Given that,

 \red{\bf :\longmapsto\:x = 7 + 4 \sqrt{3}}

So,

\rm :\longmapsto\: \sqrt{x} =  \sqrt{7 + 4 \sqrt{3} }

can be further rewritten as

\rm :\longmapsto\: \sqrt{x} =  \sqrt{4 + 3+ 4 \sqrt{3} }

\rm :\longmapsto\: \sqrt{x} =  \sqrt{ {2}^{2}  +  {( \sqrt{3} )}^{2} + 2 \times 2 \times \sqrt{3} }

\rm :\longmapsto\: \sqrt{x} =  \sqrt{ {(2 +  \sqrt{3}) }^{2} }

 \purple{\bf\implies \:\boxed{ \tt{ \:  \sqrt{x} = 2 +  \sqrt{3}}} -  -  - (1)}

Now, Consider,

\rm :\longmapsto\:\dfrac{1}{ \sqrt{x} }

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

On rationalizing the denominator, we get

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

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

\rm \:  =  \:\dfrac{2 -  \sqrt{3} }{4 - 3}

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

So,

 \purple{\bf\implies \:\boxed{ \tt{ \:   \frac{1}{ \sqrt{x} } = 2  -  \sqrt{3}}} -  -  - (2)}

Thus, On adding equation (1) and (2), we get

\rm :\longmapsto\: \sqrt{x} + \dfrac{1}{ \sqrt{x} } = 2 +  \sqrt{3} + 2 -  \sqrt{3} = 4

Hence,

 \purple{\bf\implies \:\boxed{ \tt{ \:  \sqrt{x}+\frac{1}{ \sqrt{x} } = 2  }}}

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)

Answered by XxitsmrseenuxX
1

Answer:

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

Given that,

 \red{\bf :\longmapsto\:x = 7 + 4 \sqrt{3}}

So,

\rm :\longmapsto\: \sqrt{x} =  \sqrt{7 + 4 \sqrt{3} }

can be further rewritten as

\rm :\longmapsto\: \sqrt{x} =  \sqrt{4 + 3+ 4 \sqrt{3} }

\rm :\longmapsto\: \sqrt{x} =  \sqrt{ {2}^{2}  +  {( \sqrt{3} )}^{2} + 2 \times 2 \times \sqrt{3} }

\rm :\longmapsto\: \sqrt{x} =  \sqrt{ {(2 +  \sqrt{3}) }^{2} }

 \purple{\bf\implies \:\boxed{ \tt{ \:  \sqrt{x} = 2 +  \sqrt{3}}} -  -  - (1)}

Now, Consider,

\rm :\longmapsto\:\dfrac{1}{ \sqrt{x} }

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

On rationalizing the denominator, we get

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

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

\rm \:  =  \:\dfrac{2 -  \sqrt{3} }{4 - 3}

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

So,

 \purple{\bf\implies \:\boxed{ \tt{ \:   \frac{1}{ \sqrt{x} } = 2  -  \sqrt{3}}} -  -  - (2)}

Thus, On adding equation (1) and (2), we get

\rm :\longmapsto\: \sqrt{x} + \dfrac{1}{ \sqrt{x} } = 2 +  \sqrt{3} + 2 -  \sqrt{3} = 4

Hence,

 \purple{\bf\implies \:\boxed{ \tt{ \:  \sqrt{x}+\frac{1}{ \sqrt{x} } = 2  }}}

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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