Question

if x=4-√5 then find the value of (x+1/x)²​

Answer 1

Given :-

$x = 4 - \sqrt{5}$

To find :-

$(x + \dfrac{1}{x} ) {}^{2}$

Solution :-

First we shall find value of 1/x

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

Rationalizing the denominator So, multiply with its conjugate of the denominator

Since Rationalizing factor for ,

$4 - \sqrt{5} \: \: is \: \: 4 + \sqrt{5}$

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

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

$\dfrac{1}{x} = \dfrac{4 + \sqrt{5} }{16 - 5}$

$\dfrac{1}{x} = \dfrac{4 + \sqrt{5} }{11}$

Now, finding value of x + 1/x

$x + \dfrac{1}{x} = 4 - \sqrt{5} + \dfrac{4 + \sqrt{5} }{11}$

$x + \dfrac{1}{x} = 44 - 11 \sqrt{5} + 4 + \sqrt{5} \div 11$

$x + \dfrac{1}{x} = 48 - 10 \sqrt{5} \div 11$

$(x + \dfrac{1}{x} ) {}^{2} = ( \dfrac{48 - 10 \sqrt{5} }{11} ) {}^{2}$

$(x + \dfrac{1}{x} ) {}^{2} = \dfrac{2304 + 500 -960 \sqrt{5} }{121}$

$(x + \dfrac{1}{x} ) {}^{2} = \dfrac{2804 - 960 \sqrt{5} }{121}$

Know more some algebraic identities:-

(a+ b)² = a² + b² + 2ab

( a - b )² = a² + b² - 2ab

( a + b )² + ( a - b)² = 2a² + 2b²

( a + b )² - ( a - b)² = 4ab

( a + b + c )² = a² + b² + c² + 2ab + 2bc + 2ca

a² + b² = ( a + b)² - 2ab

(a + b )³ = a³ + b³ + 3ab ( a + b)

( a - b)³ = a³ - b³ - 3ab ( a - b)

If a + b + c = 0 then a³ + b³ + c³ = 3abc

Answer 2

Basic Concept Used :-

1. Method of Rationalization :-

This method is used to remove the radicals from denominator and it means multiply and divide the given expression by conjugate of denominator.

$\begin{gathered}2. \: \: \:{\underline{\boxed{\bf{\blue{{\bf \: {(x + y)}(x - y) = {x}^{2} - {y}^{2}\: }}}}}} \\ \end{gathered}$

Given :-

$\rm :\longmapsto\:x = 4 - \sqrt{5}$

To Find :-

$\rm :\longmapsto\:\bigg(x + \dfrac{1}{x}\bigg)^{2}$

Solution :-

Given that

$\rm :\longmapsto\:x = 4 - \sqrt{5}$

So,

$\rm :\longmapsto\:\dfrac{1}{x}$

$\rm \: = \: \dfrac{1}{4 - \sqrt{5} }$

$\rm \: = \: \dfrac{1}{4 - \sqrt{5} } \times \dfrac{4 + \sqrt{5} }{4 + \sqrt{5} }$

$\rm \: = \: \dfrac{4 + \sqrt{5} }{ {4}^{2} - {( \sqrt{5})}^{2} }$

$\rm \: = \: \dfrac{4 + \sqrt{5} }{16 - 5}$

$\rm \: = \: \dfrac{4 + \sqrt{5} }{11}$

$\bf\implies \:\dfrac{1}{x} \: = \dfrac{4 + \sqrt{5} }{11}$

Now,

Consider,

$\rm :\longmapsto\:\bigg(x + \dfrac{1}{x}\bigg)^{2}$

$\rm \: = \bigg(\: 4 - \sqrt{5} + \dfrac{4 + \sqrt{5} }{11} \bigg)^{2}$

$\rm \: = \bigg(\dfrac{44 - 11\sqrt{5} + 4 + \sqrt{5} }{11} \bigg)^{2}$

$\rm \: = \bigg(\dfrac{48 - 10 \sqrt{5}}{11 }\bigg)^{2}$

$\rm \: = \: \dfrac{2304 + 500 - 960 \sqrt{5} }{121}$

$\rm \: = \: \dfrac{2804 - 960 \sqrt{5} }{121}$

Additional Information :-

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)