1.

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

Answer»

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}


Discussion

No Comment Found

Related InterviewSolutions