Question

if x=7+4√3, find √x+(1/√x)​

Answer 1

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

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)

Answer 2

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)