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Answer 1

Answer:

$\textsf{The values of x are :}$

$\dfrac{1 + i\sqrt{3} }{2}\ , \ \dfrac{1 - i\sqrt{3} }{2} \ , 2 + \sqrt{3}\ , \ 2 - \sqrt{3}$

Step-by-step explanation:

Given :

$\bigg( x^{2} + \dfrac{1}{x^2} \bigg) - 5\bigg(x + \dfrac{1}{x} \bigg) + 6 = 0, \ x \neq 0$

To Find :

The value of x

Solution :

Let us Assume,

$x + \dfrac{1}{x} = t$

Squaring on both sides

$\bigg(x + \dfrac{1}{x} \bigg)^2 = t^2$

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

$: \implies x^2 + \dfrac{1}{x^{2} } + 2(\not{x})\bigg(\dfrac{1}{\not{x}} \bigg) = t^2$

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

$: \implies x^2 + \dfrac{1}{x^{2} } = t^2 - 2 \ \ ----[1]$

So, now going back to given question,

$\bigg( x^{2} + \dfrac{1}{x^2} \bigg) - 5\bigg(x + \dfrac{1}{x} \bigg) + 6 = 0$

From [1] and our assumption,

$: \implies(t^2 - 2) -5t+6=0$

$: \implies t^2 - 5t + 4 = 0$

$: \implies t^2 - t - 4t + 4 = 0$

$: \implies t(t - 1) - 4(t-1) = 0$

$: \implies (t-1)(t-4) = 0$

$\therefore \underline{\boxed{\mathbf{ t = 1 \ (or) \ 4}}}$

But, we have assumed

$x + \dfrac{1}{x} = t$

So,

$x + \dfrac{1}{x} = 1 \ (or) \ 4$

• Case (i)

$x + \dfrac{1}{x} = 1$

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

$: \implies x^{2} - x + 1 = 0$

$\text{We know that, if a \mathrm{x^2} + bx + c = 0, then, }$

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

So, for this equation a = 1, b = -1, c = 1

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

$: \implies x = \dfrac{1 \pm \sqrt{-3} }{2}$

$: \implies x = \dfrac{1 \pm i\sqrt{3} }{2} \ \ \ [\text{as i = \sqrt{-1} }]$

$\therefore \underline{\boxed{\mathbf{ x= \dfrac{1 \pm i\sqrt{3} }{2} }}}$

• Case (ii)

$x + \dfrac{1}{x} = 4$

$: \implies x^2 + 1 = 4x$

$: \implies x^2 -4x + 1 = 0$

Here, a = 1, b = -4, c = 1

$: \implies x = \dfrac{-(-4) \pm \sqrt{16- 4(1)(1)} }{2(1)}$

$: \implies x = \dfrac{4 \pm \sqrt{12} }{2}$

$: \implies x = \dfrac{4 \pm 2\sqrt{3} }{2}$

$: \implies x = 2 \pm \sqrt{3}$

$\therefore \underline{\boxed{\mathbf{ x=2 \pm \sqrt3}}}$

Thus,

$\boxed {x = \dfrac{1 + i\sqrt{3} }{2}\ , \ \dfrac{1 - i\sqrt{3} }{2} \ , 2 + \sqrt{3}\ , \ 2 - \sqrt{3} }$

Hope it helps!!