Question

solve the
following equation:
1/x - 1/x-2 =3​

Answer 1

$\large\underline{\bold{Given \:Question - }}$

$\sf \: Solve : \: \dfrac{1}{x} - \dfrac{1}{x - 2} = 3$

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

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

$\rm :\longmapsto\: \dfrac{ \cancel{x} - 2 - \cancel{x}}{x(x - 2)} = 3$

$\rm :\longmapsto\: \dfrac{ - 2 }{x(x - 2)} = 3$

$\rm :\longmapsto\: - 2 = 3x(x - 2)$

$\rm :\longmapsto\: - 2 = 3 {x}^{2} - 6x$

$\rm :\longmapsto\: {3x}^{2} - 6x + 2 = 0$

On comparing with ax²+ bx + c = 0

we get

• a = 3

• b = - 6

• c = 2

Now,

We know that

The roots of the quadratic equation ax²+ bx + c = 0 using quadratic formula is given by

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

So, on substituting the values of a, b and c, we get

$\rm :\longmapsto\:x = \dfrac{ - ( - 6) \: \pm \: \sqrt{ {( - 6)}^{2} - 4 \times 2 \times 3 } }{2 \times 3}$

$\rm :\longmapsto\:x = \dfrac{6 \: \pm \: \sqrt{36 - 24} }{6}$

$\rm :\longmapsto\:x = \dfrac{6 \: \pm \: \sqrt{12} }{6}$

$\rm :\longmapsto\:x = \dfrac{6 \: \pm \: \sqrt{2 \times 2 \times 3} }{6}$

$\rm :\longmapsto\:x = \dfrac{6 \: \pm \: 2 \sqrt{3} }{6}$

$\rm :\longmapsto\:x = \dfrac{3 \: \pm \: \sqrt{3} }{3}$

$\bf\implies \:x = \dfrac{3 + \sqrt{3} }{3} \: \: or \: \: \dfrac{3 - \sqrt{3} }{3}$

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