Question

find the value of 1/x - 1/x-2 = 3 ​

Answer 1

Answer:

⇒ $x=1-\frac{1}{\sqrt{3}}$ (or) $x=1+\frac{1}{\sqrt{3}}$

Step-by-step explanation:

Given :

To solve for x, if

$\frac{1}{x}-\frac{1}{x-2}=3$

Solution :

$\frac{1}{x}-\frac{1}{x-2}=3$

⇒ $\frac{(x - 2) - (x)}{x(x-2)}=3$

⇒ $\frac{-2}{x^2-2x}=3$

⇒ $-2=3(x^2-2x)$

⇒ $0=3x^2-6x+2$

⇒ $0=x^2-2x+\frac{2}{3}$ (Dividing by 3)

⇒ $0=x^2-2x+\frac{2}{3}$

⇒ $0=x^2-(1+\frac{1}{\sqrt{3}})x-(1-\frac{1}{\sqrt{3}})+\frac{2}{3}$

⇒ $0=x[x-1-\frac{1}{\sqrt{3}})]-(1-\frac{1}{\sqrt{3}})(x -1-\frac{1}{\sqrt{3}})$

⇒ $0=(x-1+\frac{1}{\sqrt{3}})[x-1-\frac{1}{\sqrt{3}}]$

For the product to be equal to 0,

Either,

⇒ $x-1+\frac{1}{\sqrt{3}}=0$ (or) $x-1-\frac{1}{\sqrt{3}}=0$

⇒ $x=1-\frac{1}{\sqrt{3}}$ (or) $x=1+\frac{1}{\sqrt{3}}$

Answer 2

Answer:

$\large{\underline{\boxed{\sf x = \dfrac{3 + \sqrt{3} }{3} \: \: or \: \: \dfrac{3 - \sqrt{3} }{3} }}}$

Step-by-step explanation:

Given that-

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

Now, taking LCM of x and (x - 2), we get -

$\implies \sf \frac{1}{x} - \frac{1}{x - 2} = 3 \\ \\ \implies \sf \frac{x - 2 - x}{x(x - 2)} = 3 \\ \\ \implies \sf \frac{ - 2}{x {}^{2} - 2x} = 3 \\ \\ \bf on \: cross \: multiplying - \\ \\ \implies \sf 3(x {}^{2} - 2x) = - 2 \\ \\ \implies \sf 3x {}^{2} - 6x = - 2 \\ \\ \implies \sf 3x {}^{2} - 6x + 2 = 0$

Now, we got a quadratic equation. Solve this to get the value of x :

On comparing the given equation with the standard form of quadratic equation ax² + bx + c = 0,

a = 3

b = - 6

c = 2

Discriminant = b² - 4ac

⇒ D = (-6)² - 4 * 3 * 2

⇒ D = 36 - 24

⇒ D = 12

$\implies \tt x = \frac{ - b \pm \sqrt{D}}{2a} \\ \\ \implies \tt x = \frac{ - ( - 6) \pm \sqrt{12} }{2 \times 3} \\ \\ \implies \tt x = \frac{6 \pm 2 \sqrt{3} }{6} \\ \\ \implies \tt x = \frac{6 \pm 2 \sqrt{3} }{6} \\ \\ \implies \tt x = \frac{2(3 \pm \sqrt{3}) }{6} \\ \\ \implies \tt x = \frac{3 \pm \sqrt{3} }{3}$

Hence, the values of x are:

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