Question

please help me with this question​

Answer 1

$\large\underline{\sf{Given \:Question - }}$

Let a = 3 - sqrt(n), where n is a natural number. If p is the least positive value of a, then find the value of sqrt (p) + 1/ sqrt (p).

$\large\underline{\sf{Solution-}}$

Given that, n is a natural number and

$\rm :\longmapsto\:a = 3 - \sqrt{n}$

As p is the least value of a, so a is minimum when n is largest and keeping in view that a > 0.

So, largest possible value of n = 8.

So,

Smallest possible value of a is

$\rm :\longmapsto\:a = 3 - \sqrt{8}$

can be rewritten as

$\rm :\longmapsto\:a = 3 - \sqrt{2 \times 2 \times 2}$

$\rm :\longmapsto\:a = 3 - 2\sqrt{2}$

So,

$\rm :\longmapsto\:p = a = 3 - 2 \sqrt{2}$

$\rm :\longmapsto\:p =3 - 2 \sqrt{2}$

$\rm :\longmapsto\:p =2 + 1 - 2 \sqrt{2}$

$\rm :\longmapsto\:p = {( \sqrt{2} )}^{2} + {(1)}^{2} - 2 \times \sqrt{2} \times 1$

$\bf\implies \:p = {( \sqrt{2} - 1)}^{2}$

$\red{\bigg \{ \because \: {x}^{2} + {y}^{2} - 2xy = {(x - y)}^{2} \bigg \}}$

So,

$\bf\implies \: \sqrt{p} = \sqrt{2} - 1 - - - (1)$

Now,

$\rm :\longmapsto\:\dfrac{1}{ \sqrt{p} }$

$\rm \: = \: \dfrac{1}{ \sqrt{2} - 1}$

On rationalizing the denominator, we get

$\rm \: = \: \dfrac{1}{ \sqrt{2} - 1} \times \dfrac{ \sqrt{2} + 1}{ \sqrt{2} + 1}$

$\rm \: = \: \dfrac{ \sqrt{2} + 1}{ {( \sqrt{2} )}^{2} - {(1)}^{2} }$

$\rm \: = \: \dfrac{ \sqrt{2} + 1}{2 - 1}$

$\rm \: = \: \dfrac{ \sqrt{2} + 1}{1}$

$\bf\implies \:\dfrac{1}{ \sqrt{p} } = \sqrt{2} + 1 - - - (2)$

So, on adding equation (1) and (2), we get

$\red{\rm :\longmapsto\: \sqrt{p} + \dfrac{1}{ \sqrt{p} } = \sqrt{2} - 1 + \sqrt{2} + 1 = 2 \sqrt{2} }$

Additional Information :-

$\boxed{ \sf \: {(x + y)}^{2} = {x}^{2} + {y}^{2} + 2xy}$

$\boxed{ \sf \: {(x - y)}^{2} = {x}^{2} + {y}^{2} - 2xy}$

$\boxed{ \sf \: {(x - y)}^{3} = {x}^{3} - {y}^{3} - 3xy(x - y)}$

$\boxed{ \sf \: {(x + y)}^{3} = {x}^{3} + {y}^{3} + 3xy(x + y)}$

$\boxed{ \sf \: {x}^{2} - {y}^{2} = (x + y)(x - y)}$

$\boxed{ \sf \: {x}^{3} + {y}^{3} = (x + y)( {x}^{2} - xy + {y}^{2} )}$

$\boxed{ \sf \: {x}^{3} - {y}^{3} = (x - y)( {x}^{2} + xy + {y}^{2}) }$

$\boxed{ \sf \: {x}^{4} - {y}^{4} = (x - y)(x + y)( {x}^{2} + {y}^{2})}$

Answer 2

Answer:

$p = ( \sqrt{2} - 1) {}^{2}$