Question

if x =
$3 + 2 \sqrt{2}$
the value of
$x + \frac{1}{x}$
​

Answer 1

Answer:

6

Step-by-step explanation:

As per the provided information in the given question, we have :

$\longmapsto \rm { x = 3 + 2\sqrt{2}}$

We are asked to calculate the value of,

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

Finding the value of 1/x :

$\rm { \dfrac{1}{x}}$ is the reciprocal of x. So,

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

Rationalising the denominator. In order to the rationalise the denominator of any fraction, we multiply the rationalising factor of the denominator with both the numerator and the denominator of the fraction. Here, the denominator 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 of the fraction.

$\rm { \dfrac{1}{x} = \dfrac{1}{3+2\sqrt{2}} \times \dfrac{3 -2\sqrt{2}}{3-2\sqrt{2}} }$

Rearranging the terms.

$\rm { \dfrac{1}{x} = \dfrac{1(3 -2\sqrt{2})}{(3+2\sqrt{2})(3-2\sqrt{2})} }$

Multiplying 1 with each terms in the brackets in the numerator and using identity simplifying in the denominator of the fraction.

• (a - b)(a + b) = a² - b²

$\rm { \dfrac{1}{x} = \dfrac{3 -2\sqrt{2}}{(3)^2 - (2\sqrt{2})^2 } }$

Putting the values of the squares of the numbers in the denominator.

$\rm { \dfrac{1}{x} = \dfrac{3 -2\sqrt{2}}{9 - 8} }$

Subtracting 8 from 9.

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

Therefore, the values of the reciprocal of x is (3 - 2√2).

Now,

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

Substituting the values of x and the reciprocal of it.

$\longmapsto \rm {(3+2\sqrt{2})+(3 -2\sqrt{2}) }$

Removing the brackets.

$\longmapsto \rm {3\cancel{+2\sqrt{2}}+3 \cancel{-2\sqrt{2}} }$

Performing addition and subtraction.

$\longmapsto \bf {6}$

∴ 6 is the required answer.