Question

if x = 3+2√2, find 1) x+1/x 2) x-1/x

this is the question of 9 th standard plz answer correctly
it's to argent​

Answer 1

Answer:

$\underline{ \boxed{\sf { x + \dfrac{1}{x} = 6 }}} \; \bigstar$

$\underline{ \boxed{\sf { x - \dfrac{1}{x} = 4 \sqrt{2} }}} \; \bigstar$

Step-by-step explanation:

Here, we are given that,

• $\sf {x = 3 + 2\sqrt{2} }$

We are asked to calculate the value of :

• $\sf {x+ \dfrac{1}{x} }$

• $\sf {x- \dfrac{1}{x} }$

★ According to the question,

If $\sf {x = 3 + 2\sqrt{2} }$ , then

⇒ $\sf { \dfrac{1}{x} = \dfrac{1}{ 3 + 2\sqrt{2} } }$

Now, rationalising the denominator.

In order to rationalise the denominator, we multiply the rationalising of the denominator with both the numerator and the denominator of the fraction. Here, consider '3' as 'a' and '3 + 2√2' as 'b', so it is in the form of (a + b). Rationalising factor of (a + b) is (a - b). Therefore, rationalising factor of (3 + 2√2) is (3 - 2√2). Multiplying (3 - 2√2) with both the numerator and the denominator :-

$\longrightarrow \sf { \dfrac{1}{ 3 + 2\sqrt{2} } \times \dfrac{ 3 - 2\sqrt{2}}{ 3 - 2\sqrt{2}} }$

$\longrightarrow \sf { \dfrac{ 1(3 - 2\sqrt{2})}{ (3 + 2\sqrt{2})(3 - 2\sqrt{2})} }$

$\longrightarrow \sf { \dfrac{ 3 - 2\sqrt{2}}{ 3^2 - (2\sqrt{2})^2} }$

$\longrightarrow \sf { \dfrac{ 3 - 2\sqrt{2}}{ 9 - (4 \times 2)} }$

$\longrightarrow \sf { \dfrac{ 3 - 2\sqrt{2}}{ 9 - 8} }$

$\longrightarrow \sf { \dfrac{ 3 - 2\sqrt{2}}{ 1} }$

$\underline{ \boxed{\sf { \therefore \; \dfrac{1}{x} = 3 - 2\sqrt{2} }}}$

★ Finding the value of $\sf {x+ \dfrac{1}{x} }$ :

We have,

• $\sf {x = 3 + 2\sqrt{2} }$

• $\sf { \dfrac{1}{x} = 3 - 2\sqrt{2} }$

So,

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

$\longrightarrow \sf { (3 + 2\sqrt{2} ) + (3 - 2\sqrt{2})}$

$\longrightarrow \sf { 3 + 2\sqrt{2} + 3 - 2\sqrt{2}}$

$\longrightarrow \sf { 3 + 3 }$

$\longrightarrow \sf {6 }$

$\underline{ \boxed{\sf { \therefore \; x + \dfrac{1}{x} = 6 }}} \; \bigstar$

★ Finding the value of $\sf {x- \dfrac{1}{x} }$ :

We have,

• $\sf {x = 3 + 2\sqrt{2} }$

• $\sf { \dfrac{1}{x} = 3 - 2\sqrt{2} }$

So,

$\longrightarrow \sf {x- \dfrac{1}{x} }$

$\longrightarrow \sf { (3 + 2\sqrt{2} ) - (3 - 2\sqrt{2})}$

$\longrightarrow \sf { 3 + 2\sqrt{2} - 3 + 2\sqrt{2}}$

$\longrightarrow \sf {2\sqrt{2} + 2\sqrt{2} }$

$\longrightarrow \sf {(2+2)\sqrt{2} }$

$\longrightarrow \sf {4 \sqrt{2} }$

$\underline{ \boxed{\sf { \therefore \; x - \dfrac{1}{x} = 4 \sqrt{2} }}} \; \bigstar$

_______________________________

Points to remember :

• Rationalising factor of (a + b) is (a - b)

• (a + b)² = a² + b² + 2ab

• (a - b)² = a² + b² - 2ab

• (a - b)(a + b) = a² - b²

• $\sf { { \Bigg \lgroup x + \dfrac{1}{x} \Bigg \rgroup}^{2} = x^2 + \dfrac{1}{x^2} + 2}$

• $\sf { { \Bigg \lgroup x - \dfrac{1}{x} \Bigg \rgroup}^{2} = x^2 + \dfrac{1}{x^2} - 2}$