Question

if x=9 + 4root5, then find the value of rootx - 1 by root x.​

Answer 1

GIVEN :-

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

TO FIND :-

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

SOLUTION :-

We have ,

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

Taking reciprocal,

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

Rationalising the denominator,

Rationalising factor is (9 - 4√5)

$\\ \implies\sf \: \dfrac{1}{x} = \dfrac{1}{9 + 4\sqrt{5} } \times \dfrac{9 - 4\sqrt{5} }{9 - 4\sqrt{5} } \\ \\ \implies\sf \: \dfrac{1}{x} = \dfrac{9 - 4\sqrt{5} }{(9 + 4\sqrt{5})(9 - 4\sqrt{5} ) }$

In denominator,

★ (a+b)(a-b) = a² - b²

• a = 9
• b = 4√5

Putting values,

$\\ \implies\sf \: \dfrac{1}{x} = \dfrac{9 - 4\sqrt{5} }{ {9}^{2} - {( 4\sqrt{5} )}^{2} } \\ \\ \implies\sf \: \dfrac{1}{x} = \dfrac{9 - 4\sqrt{5} }{81 - 80} \\ \\ \implies\sf \: \dfrac{1}{x} = \dfrac{9 - 4\sqrt{5} }{1} \\ \\ \implies\underline{ \sf \: \dfrac{1}{x} = 9 - 4\sqrt{5} }$

Now ,

$\\ \sf \: Let \: \: \: \: y = \sqrt{x} - \dfrac{1}{ \sqrt{x} }$

Squaring both sides,

$\\ \implies\sf \: {y}^{2} = {\left( \sqrt{x} - \dfrac{1}{ \sqrt{x} } \right)}^{2}$

★ (a-b)² = a² + b² - 2ab

• a = √x
• b = 1/√x

Putting values,

$\\ \implies\sf \: {y}^{2} = { (\sqrt{x} )}^{2} + {\left( \dfrac{1}{ \sqrt{x} } \right)}^{2} - 2( \cancel{ \sqrt{x}} )\left( \dfrac{1}{ \cancel{\sqrt{x} }} \right) \\ \\ \implies\sf \: {y}^{2} = x + \dfrac{1}{x} - 2$

Putting values of x and 1/x..

$\\ \implies\sf \: {y}^{2} = 9 + \cancel{4 \sqrt{5}} + 9 - \cancel{4 \sqrt{5}} - 2 \\ \\ \implies\sf \: {y}^{2} = 16 \\ \\ \implies\boxed{\sf \: y = 4 \: , \: - 4}$

Hence ,

$\\ \boxed{\bf \: \sqrt{x} - \dfrac{1}{ \sqrt{x} } = 4 \: , - 4}$