Question

(x²+(1/x²)) - 5(x-(1/x)) +2 = 0​

Answer 1

$\large\underline{\sf{Solution-}}$

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

$\red{\rm :\longmapsto\:Let \: x - \dfrac{1}{x} = y} \\ \blue{\rm \: on \: squaring \: both \: sides} \\ \red{\rm \: {x}^{2} + \dfrac{1}{ {x}^{2}} - 2 = {y}^{2} }\\ \red{\rm \: {x}^{2} + \dfrac{1}{ {x}^{2} } = {y}^{2} + 2}$

So,

Equation (1) can be rewritten as

$\rm :\longmapsto\: {y}^{2} + 2 - 5y + 2 = 0$

$\rm :\longmapsto\: {y}^{2} - 5y + 4 = 0$

$\rm :\longmapsto\: {y}^{2} - 4y - y + 4 = 0$

$\rm :\longmapsto\:y(y - 4) - 1(y - 4) = 0$

$\rm :\longmapsto\:(y - 4)(y - 1) = 0$

$\bf\implies \:y = 1 \: \: \: or \: \: \: y = 4$

Case :- 1

When y = 1

$\rm :\longmapsto\:x - \dfrac{1}{x} = 1$

$\rm :\longmapsto\:\dfrac{ {x}^{2} - 1 }{x} = 1$

$\rm :\longmapsto\: {x}^{2} - 1 = x$

$\rm :\longmapsto\: {x}^{2} - x - 1 = 0$

Using quadratic formula,

We know,

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

Here,

• a = 1

• b = - 1

• c = - 1

On substituting all these values in above formula, we get

$\rm :\longmapsto\:x = \dfrac{ - ( - 1) \: \pm \: \sqrt{ {( - 1)}^{2} - 4 \times 1 \times ( - 1)}}{2 \times 1}$

$\rm :\longmapsto\:x = \dfrac{1\: \pm \: \sqrt{1 + 4}}{2}$

$\rm :\longmapsto\:x = \dfrac{1\: \pm \: \sqrt{5}}{2}$

$\bf\implies \:x = \dfrac{1 \: + \: \sqrt{5} }{2} \: \: \: or \: \: \: \dfrac{1 \: - \: \sqrt{5} }{2}$

Case :- 2

When y = 4

$\rm :\longmapsto\:x - \dfrac{1}{x} = 4$

$\rm :\longmapsto\:\dfrac{ {x}^{2} - 1 }{x} = 4$

$\rm :\longmapsto\: {x}^{2} - 1 = 4x$

$\rm :\longmapsto\: {x}^{2} - 4x - 1 = 0$

Again, using quadratic formula,

We know,

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

Here,

• a = 1

• b = - 4

• c = - 1

On substituting all these values in above formula, we have

$\rm :\longmapsto\:x = \dfrac{ - ( - 4) \: \pm \: \sqrt{ {( - 4)}^{2} - 4 \times 1 \times ( - 1)}}{2 \times 1}$

$\rm :\longmapsto\:x = \dfrac{4\: \pm \: \sqrt{16 + 4}}{2}$

$\rm :\longmapsto\:x = \dfrac{4\: \pm \: \sqrt{20}}{2}$

$\rm :\longmapsto\:x = \dfrac{4\: \pm \: \sqrt{2 \times 2 \times 5}}{2}$

$\rm :\longmapsto\:x = \dfrac{4\: \pm \: 2\sqrt{5}}{2}$

$\rm :\longmapsto\:x = 2 \: \pm \: \sqrt{5}$

$\bf\implies \:x = 2 + \sqrt{5} \: \: \: or \: \: \: 2 - \sqrt{5}$

Hence,

$\bf\implies\:x=\dfrac{1+\sqrt{5} }{2} \: or \:\dfrac{1-\sqrt{5} }{2}\:or\:2+ \sqrt{5}\:or\:2-\sqrt{5}$

Additional Information :-

Nature of roots :-

Let us consider a quadratic equation ax² + bx + c = 0, then nature of roots of quadratic equation depends upon Discriminant (D) of the quadratic equation.

• If Discriminant, D > 0, then roots of the equation are real and unequal.

• If Discriminant, D = 0, then roots of the equation are real and equal.

• If Discriminant, D < 0, then roots of the equation are unreal or complex or imaginary.

Where,

• Discriminant, D = b² - 4ac