Question

Bello Brainliest users!
Math : Please be detailed
Question 1 :
If x =
$\sf \dfrac{1}{3 - \sqrt{5} }$
Find :
$\sf \sqrt{x} + \dfrac{1}{ \sqrt{x} }$
Question 2 :
Let a = 3 - √n, where n is a natural number. If 'p' is the least possible value of 'a', then find the value of :
$\sf \sqrt{p} + \dfrac{1}{ \sqrt{p} }$

​

Answer 1

Question 1:

It is given that:

$\sf{\implies\:x=\dfrac{1}{3-\sqrt{5}}}$

And we are asked to find:

$\sf{\implies\:\sqrt{x}+\dfrac{1}{\sqrt{x}}}$

Solving:

✳ First we have to rationalize the given value of the variable x.

$\sf{\longrightarrow\:\dfrac{1}{3-\sqrt{5}}\times\dfrac{3+\sqrt{5}}{3+\sqrt{5}}}$

✳ The result will be:

$\sf{\longrightarrow\:\dfrac{3+\sqrt{5}}{9-5}}$

$\sf{\longrightarrow\:\dfrac{3+\sqrt{5}}{4}}$

This is the simplified value of variable 'x'.

✳ This means that:

$\sf{\implies\:\dfrac{1}{x}=3-\sqrt{5}}$

Now, lets consider the following statement:

$\sf{\implies\:\bigg(\sqrt{x}}+\dfrac{1}{\sqrt{x}}\bigg)^2=x+\dfrac{1}{x}+2$

✳ Here let's consider the RHS of the equation and substitute values based on it:

$\sf{\longrightarrow\:x+\dfrac{1}{x}+2}$

$\sf{\longrightarrow\:\dfrac{3+\sqrt{5}}{4}+3-\sqrt{5}+2}$

✳ Now moving the complete expression under the denominator 4:

$\sf{\longrightarrow\:\dfrac{3+\sqrt{5}+12-4\sqrt{5}+8}{4}}$

✳ Simplifying the expression further, we get:

$\sf{\longrightarrow\:\dfrac{23-3\sqrt{5}}{4}}$

✳ Multiplying both the numerator and denominator with 2 (for further simplifications):

$\sf{\longrightarrow\:\dfrac{46-6\sqrt{5}}{8}}$

✳ Expanding the numerator part of the fraction (for further simplifications):

$\sf{\longrightarrow\:\dfrac{45+1-6\sqrt{5}}{8}}$

Now the numerator is the format:

➡ (a - b)² = a² - 2ab + b²

✳ Therefore, converting the numerator part into the given identity format:

$\sf{\longrightarrow\:\bigg(\dfrac{3\sqrt5-1}{\sqrt{8}}\bigg)^2}$

✳ The denominator part can be written as:

$\sf{\longrightarrow\:\bigg(\dfrac{3\sqrt5-1}{2\sqrt{2}}\bigg)^2}$

✳ The given expression is in the format:

$\sf{\implies\:\bigg(\sqrt{x}}+\dfrac{1}{\sqrt{x}}\bigg)^2$

✳ This means that:

$\sf{\bigg(\sqrt{x}}+\dfrac{1}{\sqrt{x}}\bigg)^2=\bigg(\dfrac{3\sqrt5-1}{2\sqrt{2}}\bigg)^2$

Henceforth:

$\bf{\sqrt{x}}+\dfrac{1}{\sqrt{x}}=\dfrac{3\sqrt5-1}{2\sqrt{2}}$

Question 2:

It is given that:

➡ a = 3 - √n

✳ If we give the value of n as 9, then the value of a will become 0 which will be a whole number. [√9 is 3 and 3 - 3 = 0]

Therefore, the value of "n" must be taken as 8 [ n = 8 ].

This means that the least possible value "p" for a is:

$\sf{\longrightarrow\:p=3-\sqrt{8}}$

So,

$\sf{\longrightarrow\:\sqrt{p}=\sqrt{3-\sqrt8}}$

$\sf{=\sqrt{3-2\sqrt2}}$

$\sf{=\sqrt{\sqrt2-1}}$

$\sf{=\sqrt{2}-1}$

✳ Now,

$\sf{\longrightarrow\:\dfrac{1}{\sqrt{p}}=\dfrac{1}{\sqrt2-1}}$

✳ Rationalizing the denominator:

$\sf{=\dfrac{1}{\sqrt2-1}\times\dfrac{\sqrt2+1}{\sqrt2+1}}$

$\sf{=\sqrt{2}+1}$

So,

$\bf{\star\:\sqrt{p}=\sqrt{2}-1}$

$\bf{\star\:\dfrac{1}{\sqrt{p}}=\sqrt{2}+1}$

✳ Now, we have to find:

$\sf{\longrightarrow\:\sqrt{p}+\dfrac{1}{\sqrt{p}}}$

✳ Substituting the values we found into the expression:

$\sf{\longrightarrow\:\sqrt2-1+\sqrt2+1}$

$\bf{=2\sqrt{2}}$

Henceforth:

$\bf{\sqrt{p}+\dfrac{1}{\sqrt{p}}=2\sqrt2}$