Question

If
$x$
=
$9 + 4 \sqrt{5}$
find
$\sqrt{x} - \frac{1}{ \sqrt{x} }$
​

Answer 1

Given :-

x = 9 + 4√5

To find :-

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

Solution :-

We know that :- (a-b)² = a²+b²-2ab

$\rm\therefore \rm {\bigg(\rm \sqrt{x} - \dfrac{1}{ \sqrt{x} } \bigg)}^{2} = {( \sqrt{x}) }^{2} + { \bigg(\dfrac{1}{ \sqrt{x} } \bigg)}^{2} - \bigg( 2 \times \sqrt{x} \times \dfrac{1}{ \sqrt{x}} \bigg)$

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

Now, value of x is given but we have to find value of (1/x).

x = 9 + 4√5

$\rm \rightarrow \dfrac{1}{x} = \dfrac{1}{9 + 4 \sqrt{5} }$

$\rm \rightarrow \dfrac{1}{x} = \dfrac{1}{9 + 4 \sqrt{5} } \times\dfrac{9 - 4 \sqrt{5}}{9 - 4 \sqrt{5} }$

$\rm \rightarrow \dfrac{1}{x} = \dfrac{9 - 4 \sqrt{5}}{ {(9)}^{2} - {(4 \sqrt{5})}^{2} }$

$\rm \rightarrow \dfrac{1}{x} = \dfrac{9 - 4 \sqrt{5}}{ 81 - 80 }$

$\rm \rightarrow \dfrac{1}{x} = \dfrac{9 - 4 \sqrt{5}}{ 1 }$

$\rm \rightarrow \dfrac{1}{x} = {9 - 4 \sqrt{5}}$

$\rm\longrightarrow \rm {\bigg(\rm \sqrt{x} - \dfrac{1}{\sqrt{x} } \bigg)}^{2} = 9 + 4 \sqrt{5} + 9 - 4 \sqrt{5} - 2$

$\rm\longrightarrow \rm {\bigg(\rm \sqrt{x} - \dfrac{1}{\sqrt{x} } \bigg)}^{2} = 18 - 2$

$\rm\longrightarrow \rm {\rm \sqrt{x} - \dfrac{1}{\sqrt{x} } } = \sqrt{16}$

$\rm\longrightarrow \rm {\rm \sqrt{x} - \dfrac{1}{\sqrt{x} } } = \pm4$

Answer : ± 4

Answer 2

$\sf \Large Solution!!$

$\sf \large Given:$

$\sf \to x=9+4\sqrt{5}$

$\sf \large To\:find:$

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

We will use the identity: (a-b)² = a² + b² - 2ab

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

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

Now, we have to find the value of $\sf \dfrac{1}{x}$

$\sf \to \dfrac{1}{x}=\dfrac{1}{9+4\sqrt{5}}$

Rationalising the denominator.

$\sf \to \quad =\dfrac{1}{9+4\sqrt{5}}\times \dfrac{9-4\sqrt{5}}{9-4\sqrt{5}}$

$\sf \to \quad =\dfrac{9-4\sqrt{5}}{(9+4\sqrt{5})(9-4\sqrt{5})}$

We can use the identity → (a+b)(a-b)=a² - b² ← in the denominator.

$\sf \to \quad =\dfrac{9-4\sqrt{5}}{(9)^{2}-(4\sqrt{5})^{2}}$

$\sf \to \quad =\dfrac{9-4\sqrt{5}}{81-80}$

$\sf \to \quad =\dfrac{9-4\sqrt{5}}{1}$

$\sf \to \dfrac{1}{x}=9-4\sqrt{5}$

Now, let's apply these values in the formula $\\→ \bigg(\sqrt{x}-\dfrac{1}{\sqrt{x}}\bigg)^{2}=x+\dfrac{1}{x}-2 ←$

$\sf \to \bigg(\sqrt{x}-\dfrac{1}{\sqrt{x}}\bigg)^{2}=(9+4\sqrt{5})+(9-4\sqrt{5})-2$

$\sf \to \bigg(\sqrt{x}-\dfrac{1}{\sqrt{x}}\bigg)^{2}=9\cancel{+4\sqrt{5}}+9\cancel{-4\sqrt{5}}-2$

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

$\sf \to \sqrt{x}-\dfrac{1}{\sqrt{x}}=\sqrt{16}$

$\bold{\boxed{\blue{\sf \sqrt{x}-\dfrac{1}{\sqrt{x}}=±4}}}$