Question

if x=7-4√3 then find √x+1/√x​

Answer 1

» Question :

If $\sf{x = 7 - 4\sqrt{3}}$ then ,find the value of $\sf{\sqrt{x} - \dfrac{1}{\sqrt{x}}}$

» To Find :

The value of $\sf{\sqrt{x} - \dfrac{1}{\sqrt{x}}}$

» Given :

• $\sf{x = 7 - 4\sqrt{3}}$

» We Know :

• $\sf{\left(x + \dfrac{1}{x}\right)^{2} = x^{2} + \left(\dfrac{1}{x}\right)^{2} + 2}$

• $\sf{a^{2} - b^{2} = (a + b)(a - b)}$

» Concept :

According to the mathematics , if $\sf{x = 7 - 4\sqrt{3}}$ ,then $\sf{\dfrac{1}{x} = 7 + 4\sqrt{3}}$

Proof :

Given :

$\sf{x = 7 - 4\sqrt{3}}$(Equation...(i))

Then :

$\sf{\dfrac{1}{x} = \dfrac{1}{7 - 4\sqrt{3}}}$

Multiplying the Equation by $\sf{7 + 4\sqrt{3}}$ on both the numerator and denominator.

$\sf{\Rightarrow \dfrac{1}{x} = \dfrac{1}{7 - 4\sqrt{3}} \times \dfrac{7 + 4\sqrt{3}}{7 + 4\sqrt{3}}}$

$\sf{\Rightarrow \dfrac{1}{x} = \dfrac{7 + 4\sqrt{3}}{7 - 4\sqrt{3} \times 7 + 4\sqrt{3}}}$

Using the identity :

$\sf{a^{2} - b^{2} = (a + b)(a - b)}$

we get :

$\sf{\Rightarrow \dfrac{1}{x} = \dfrac{7 + 4\sqrt{3}}{7^{2} - \bigg(4\sqrt{3}\bigg)^{2}}}$

$\sf{\Rightarrow \dfrac{1}{x} = \dfrac{7 + 4\sqrt{3}}{49 - 48}}$

$\sf{\Rightarrow \dfrac{1}{x} = \dfrac{7 + 4\sqrt{3}}{1}}$

$\sf{\Rightarrow \dfrac{1}{x} = 7 + 4\sqrt{3}}$

$\sf{\therefore \dfrac{1}{x} = 7 + 4\sqrt{3}}$(Equation...(ii))

Now ,by this information we can find the required value .

» Solution :

Equation (i)

$\sf{x = 7 - 4\sqrt{3}}$

Equation (ii)

$\sf{\therefore \dfrac{1}{x} = 7 + 4\sqrt{3}}$

On adding the two Equations ,we get :

$\sf{x + \dfrac{1}{x}}$

$\sf{\Rightarrow 7 - 4\sqrt{3} + 7 + 4\sqrt{3}}$

$\sf{\Rightarrow 7 - \cancel{4\sqrt{3}} + 7 + \cancel{4\sqrt{3}}}$

$\sf{\Rightarrow 7 + 7}$

$\sf{\Rightarrow 14}$

Hence , $\sf{x + \dfrac{1}{x} = 14}$

Given Equation to find :

$\sf{\sqrt{x} - \dfrac{1}{\sqrt{x}}}$

Squaring the equation,we get :

$\sf{\bigg(\sqrt{x} - \dfrac{1}{\sqrt{x}}\bigg)^{2}}$

$\sf{\Rightarrow \bigg(\sqrt{x} - \dfrac{1}{\sqrt{x}}\bigg)^{2} = x + \dfrac{1}{x} + 2}$

Putting the value of $\sf{x + \dfrac{1}{x} = 14}$ ,in the Equation ,we get :

$\sf{\Rightarrow \bigg(\sqrt{x} - \dfrac{1}{\sqrt{x}}\bigg)^{2} = 14 + 2}$

$\sf{\Rightarrow \bigg(\sqrt{x} - \dfrac{1}{\sqrt{x}}\bigg)^{2} = 16}$

On Square rooting on both the sides ,we get ;

$\sf{\Rightarrow \sqrt{\bigg(\sqrt{x} - \dfrac{1}{\sqrt{x}}\bigg)^{2}} = \sqrt{16}}$

$\sf{\Rightarrow \bigg(\sqrt{x} - \dfrac{1}{\sqrt{x}}\bigg) = 4}$

Hence ,the value of $\sf{\sqrt{x} - \dfrac{1}{\sqrt{x}}}$ is 4.

» Additional information :

• (a² + b²) = (a + b)² - 2ab

• (a² + b²) = (a - b)² + 2ab

• (a + b)² = a² + b² + ab

• a³ + b³ = (a + b)(a² - 2ab + b²)

• a³ - b³ = (a - b)(a² + 2ab + b²)

Answer 2

Given :

$\rightarrow \mathtt{x = 7 - 4 \sqrt{3} }$

To find :

$\rightarrow \mathtt{ \sqrt{x} + \dfrac{1}{ \sqrt{x} } }$

Solution :

As per the given question ,

$\rightarrow \mathtt{x = 7 - 4 \sqrt{3} }$

This equation can also be written as ,

$\rightarrow \mathtt{x = 4 + 3 - 4 \sqrt{3} }$

$\rightarrow \mathtt{x = {(2)}^{2} - 2 \times 2 \sqrt{3} + {( \sqrt{3}) }^{2} }$

we know that ,

$\star \boxed{\mathtt{ {(a - b)}^{2} = {a}^{2} - 2ab + {b}^{2} }}$

Hence ,

$\rightarrow \mathtt{x = {(2 - \sqrt{3}) }^{2} }$

$\rightarrow \mathtt{ \sqrt{x} = \sqrt{( {2 - \sqrt{3} )}^{2} } }$

$\rightarrow \mathtt{ \pink{ \sqrt{x} = 2 - \sqrt{3} }}$

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$\rightarrow \mathtt{ \sqrt{x} + \dfrac{1}{ \sqrt{x} } }$

Substituting the value of √ x in the above equation we get ,

$\mathtt{ = 2 - \sqrt{3} + \dfrac{1}{2 - \sqrt{3} } }$

$\mathtt{ = \dfrac {{(2 - \sqrt{3} )}^{2} + 1}{2 - \sqrt{3} }}$

$\mathtt{ = \dfrac {7 - 4 \sqrt{3} + 1}{2 - \sqrt{3} }}$

$\mathtt{ = \dfrac {8- 4 \sqrt{3} }{2 - \sqrt{3} }}$

$\mathtt{ = \dfrac {4 \cancel{(2- \sqrt{3} )} }{ \cancel{2 - \sqrt{3}} }}$

hence ,

$\rightarrow \mathtt{ \red{ \sqrt{x} + \dfrac{1}{ \sqrt{x} } = 4}}$

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Hope this helps :D