Question

17. If x + 1 = 6, find (i) x-1/x (ii) x^2 - 1/x^2
​(Answer of (i) is -+4√3 and answer of (ii) is -+24√2)

Answer 1

Answer :-

◾Given :-

▪️$x + \dfrac{1}{x} = 6$

$\implies \dfrac{x^2 + 1}{x} = 6$

$\implies x^2 + 1 = 6x$

$\implies x^2 - 6x + 1 = 0$

» Value of x by using

$\dfrac{-b \pm \sqrt{ b^2 - 4ac}}{2a}$

$= \dfrac{-(-6) \pm \sqrt{ (-6)^2 - 4(1)(1)}}{2}$

$= \dfrac{ 6 \pm \sqrt{ 36 - 4}}{2}$

$= \dfrac{ 6 \pm \sqrt{ 32}}{2}$

$= \dfrac{ 6 \pm 4\sqrt {2 }}{2}$

$= 3 \pm 2\sqrt{2}$

So value of x

$= 3 + 2\sqrt{2}$

or

$= 3 - 2\sqrt{2}$

**Note :- When we rationalize 1/x it will be in the form of other root .

◾Now coming to the question :-

(i) $x - \dfrac{1}{x}$

:- When x $= 3 + 2\sqrt{2}$

Then by using above's note .

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

$= 3 + 2\sqrt{2} -(3 - 2\sqrt{2})$

$= 3 + 2\sqrt{2} - 3 + 2\sqrt{2}$

$= 4\sqrt{2}$

:- When x $= 3 + 2\sqrt{2}$

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

$= 3 - 2\sqrt{2} -(3 + 2\sqrt{2})$

$= 32\sqrt{2} - 3 - 2\sqrt{2}$

$= -4\sqrt{2}$

(ii) $x^2 - \dfrac{1}{x^2}$

▪️As $a^2 - b^2 = (a-b)(a+b)$

Then when

:- $x + \dfrac{1}{x} = 4\sqrt{2}$

$\implies x^2 - \dfrac{1}{x^2} = (4\sqrt{2})(6)$

$\implies x^2 - \dfrac{1}{x^2} = 24\sqrt{2}$

:- $x + \dfrac{1}{x} = -4\sqrt{2}$

$\implies x^2 - \dfrac{1}{x^2} = (-4\sqrt{2})(6)$

$\implies x^2 - \dfrac{1}{x^2} = -24\sqrt{2}$

⏩So

$\bold{x - \dfrac{1}{x} =\pm 4\sqrt{2}}$

$\bold{ x^2 - \dfrac{1}{x^2} =\pm 24\sqrt{2}}$