Question

if x=2+root 3 find the value of x square-1/x square​

Answer 1

Step-by-step explanation:

Here is your solution

Given :-

x=2+√3

Now

\begin{gathered} \frac{1}{x} = \frac{1}{2 +\sqrt{3} } \times \frac{2 -\sqrt{3} }{2 - \sqrt{3} } \\ \frac{1}{x} = \frac{2 + \sqrt{3} }{2 {}^{2} - ( \sqrt{3}) {}^{2} } \\ \frac{1}{x} = \frac{2 + \sqrt{3} }{4 - 3} \\ \frac{1}{x} = 2 - \sqrt{3} \\ \\ \\ \end{gathered}

x

1

=

2+

3

1

×

2−

3

2−

3

x

1

=

2

2

−(

3

)

2

2+

3

x

1

=

4−3

2+

3

x

1

=2−

3

\begin{gathered} x + \frac{1}{x} = 2 - \sqrt{3} + 2 + \sqrt{3} \\ x + \frac{1}{x} = 4 \\ Both \: sides \: squaring. \: \\ (x + \frac{1}{x} ) {}^{2} = 4 {}^{2} \\ x {}^{2} + \frac{1}{x {}^{2} } + 2 = 16 \\ x {}^{2} + \frac{1}{x {}^{2} } = 16 - 2 \\ x {}^{2} + \frac{1}{x {}^{2} } = 14\end{gathered}

x+

x

1

=2−

3

+2+

3

x+

x

1

=4

Bothsidessquaring.

(x+

x

1

)

2

=4

2

x

2

+

x

2

1

+2=16

x

2

+

x

2

1

=16−2

x

2

+

x

2

1

=14

Answer 2

$8 \sqrt{3}$

Step-by-step explanation:

$x = 2 + \sqrt{3}$

$\frac{1}{x} = \frac{1}{2 + \sqrt{3} }$

On rationalising,

$\frac{1}{x} = 2 - \sqrt{3}$

As we know,

$x ^{2} - y ^{2} = (x - y)(x + y)$

Using this,

$x ^{2} - \frac{1}{x ^{2} } = (x - \frac{1}{x} )( x + \frac{1}{x} )$

So putting values of x and 1/x, we get:

$x ^{2} - \frac{1}{ {x}^{2} } = (2 + \sqrt{3} - (2 - \sqrt{3} )) (2 + \sqrt{3} + 2 - \sqrt{3} )$

Then,

$x ^{2} - \frac{1}{ {x}^{2} } = 4 \times 2 \sqrt{3} = 8 \sqrt{3}$